BERKELEY 

LIBRARY 

UNIVERSITY  Of 
CAIIPORNIA 


A  SYSTEMATIC  QUALITATIVE 
CHEMICAL  ANALYSIS 

A  Theoretical  and  Practical  Study  of 
Analytical  Reactions  of  the  More 
Common  Ions  of  Inorganic  Substances 


BY 

GEO.  W.  SEARS,  Ph.D. 

ASSOCIATE  PROFESSOR  OF  CHEMISTRY  IN  THE 
UNIVERSITY  OF  NEVADA 


NEW    YORK 

JOHN  WILEY  &  SONS,  Inc. 

LONDON:  CHAPMAN  &  HALL,  LIMITED 
1922 


Copyright,  1922 
BY  GEORGE  W.  SEARS 


LOAN  STAO5 


PRESS  OF 

BRAUNWORTH   a  CO. 

BOOK   MANUFACTURERS 

BROOKLYN,   N.  Y. 


QP  8 1 


PREFACE 


THE  author  believes  that  a  course  in  Qualitative  Analysis 
should  not  only  train  the  student  in  accurate  and  careful 
manipulation  and  so  prepare  him  for  the  more  careful 
work  necessary  in  Quantitative  Analysis,  but  should  also 
serve  to  extend  and  supplement  his  knowledge  of  General 
and  Theoretical  Chemistry.  To  this  end  an  attempt  has 
been  made  to  present  the  theoretical  and  practical  parts 
in  such  a  way  that  the  student  will  understand  the  signifi- 
cance and  see  the  practical  applications  of  the  theoretical 
part. 

The  book  has  been  divided  into  four  parts,  including 
the  Appendix.  Part  I  consists  of  the  Introduction  and 
Laboratory  Suggestions.  In  the  Introduction  is  given  a 
brief  discussion  of  the  Ionic  Theory  and  Law  of  Mass 
Action,  as  applied  to  the  principles  involved  in  Qualitative 
Analysis,  and  reference  to  its  various  sections  is  made 
throughout  the  text.  Part  II  consists  of  Preliminary 
Experiments,  Method  of  Analysis,  Discussion  and  Review 
Questions  on  the  metal  ions.  In  -order  that  the  student 
may  get  a  comparative  idea  of  the  actions  of  the  different 
metal  ions  in  a  group  toward  a  given  reagent,  the  Pre- 
liminary Experiments  are  arranged  in  a  manner  which 
differs  somewhat  from  the  usual  one  and  in  order  that 
he  may  not  be  confused  by  too  many  reactions,  only  those 
reagents  used  in  the  Method  of  Analysis  are  employed. 
Preliminary  Experiments  whose  reactions  may  not  be  clear 
are  followed  by  Notes.  In  the  Method  of  Analysis  an 
attempt  has  been  made  to  give  clear  and  concise  directions 
for  procedure  only,  all  explanations  and  conditional  pro- 

iii 


[      2C1 


iv  PREFACE 

cedures  being  placed  together  under  Discussion,  following 
each  group  or  sub-group.  It  is  believed  that  by  frequent 
reference  to  this  the  student's  attention  will  be  brought 
more  effectively  to  the  theoretical  explanations  and  to 
the  reasons  for  careful  work.  In  so  far  as  practicable, 
only  Quantitative  reactions  have  been  given  and  only 
those  methods  have  been  used  which  have  been  thoroughly 
tested.  Part  III  consists  of  Method  of  Analysis,  Dis- 
cussion and  Review  Questions  on  the  acids. 

A  new  method  of  anion  analysis  is  given  which  follows 
in  general  the  method  of  procedure  in  metal  analysis,  in 
that  a  single  sample  is  taken  for  analysis,  and  largely  by 
means  of  precipitation  methods,  the  separadon  and  detec- 
tion of  anions  are  made.  Many  of  the  same  reactions 
used  in  the  course  of  metal  analysis  are  also  employed. 
For  these  reasons  no  preliminary  experiments  are  given, 
the  author  having  found  that  students  are  able  to  under- 
stand and  follow  the  directions  in  the  analysis  of 
"  unknowns."  In  the  Appendix  are  given  directions  for 
the  preparation  of  reagents  and  test  solutions,  also  tables  of 
solubilities  and  atomic  weights  of  the  more  common 
elements. 

In  the  preparation  of  this  text  the  author  has  made 
use  of  information  from  all  convenient  sources,  including 
such  textbooks  on  Qualitative  Analysis  as  those  of  Tread- 
well-Hall,  A.  A.  Noyes,  Julius  Stieglitz,  and  W.  A.  Noyes, 
to  whom  he  wishes  to  acknowledge  his  indebtedness. 
Acknowledgment  is  also  due  to  Dr.  J.  F.  G.  Hicks,  Stan- 
ford University,  and  to  Miss  M.  Dewar,  University  of 
Nevada,  for  helpful  criticism  and  assistance. 


CONTENTS 


PART  I 

PAGE 

Introduction I 

Laboratory  Suggestions 19 

PART  II 

The  Systematic  Analysis  (Cations) 22 

Preparation  of  Solution 22 

Discussion  (l-io) 26 

Group  I. 

Preliminary  Experiments 29 

Outline  of  Analysis  (Table  I) 31 

Analysis 31 

Discussion  (11-15) 32 

Group  II. 

Preliminary  Experiments  (Cu  division) 34 

Outline  of  Analysis  (Table  II) 38 

Analysis  (Separation  into  Cu  and  Sn  divisions) 38 

Discussion  (16-20) 39 

Outline  of  Analysis  (Table  III) 42 

Analysis  (Cu  division) 43 

Discussion  (21-27) 44 

Preliminary  Experiments  (Sn  division) 46 

Outline  of  Analysis  (Table  IV) 50 

Analysis  (Sn  division) 50 

Discussion  (28-31) 52 

Group  III. 

Preliminary  Experiments  (Al  division) 53 

Outline  of  Analysis  (Table  V) 56 

Analysis  (Separation  into  Al  and  Fe  divisions) 56 

Discussion  (32-37) 57 

Outline  of  Analysis  (Table  VI) 60 

Analysis  (Al  division) 60 

Discussion  (38-41) 62 

Preliminary  Experiments  (Fe  division) 63 

Outline  of  Analysis  (Table  VII) 66 

Outline  of  Analysis  (Table  VIII) 66 

v 


vi  CONTENTS 

PAGE 

Analysis  (Fe  division) 67 

Discussion  (42-48) 70 

Group  IV. 

Preliminary  Experiments 73 

Outline  of  Analysis  (Table  IX) 75 

Analysis 75 

Discussion  (49-53) 77 

Group  V. 

Preliminary  Experiments 79 

Outline  of  Analysis  (Table  X) 81 

Analysis 81 

Discussion  (54-58) 82 

Questions  for  Review 84 

PART  III 

Acids  (General  statement) 88 

The  Systematic  Analysis  (Anions) 89 

Preparation  of  Solution 89 

Discussion  (59-60) 90 

Group  I. 

Outline  of  Analysis,  Division  A  (Table  XI) 91 

Analysis  (Division  A) 91 

Discussion  (61-63) 92 

Outline  of  Analysis,  Division  B  (Table  XII) 94 

Analysis  (Division  B) 94 

Discussion  (64-66) 96 

Group  II. 

Outline  of  Analysis  (Table  XIII) 98 

Analysis 98 

Discussion  (67-69) 99 

Group  III. 

Outline  of  Analysis  (Table  XIV) 102 

Analysis 103 

Discussion  (70-75) 105 

Group  IV. 

Outline  of  Analysis  (Table  XV) 107 

Analysis , 107 

Discussion  (76-80) 109 

Questions  for  Review 1 1 1 

APPENDIX 

I.  Preparation  of  Reagents 113 

II.  Preparation  of  Test  Solutions 1 16 

III.  Table  of  Solubilities 118 

IV.  Table  of  Atomic  Weights 119 


A  SYSTEMATIC 
QUALITATIVE  CHEMICAL  ANALYSIS 


PART  I 


INTRODUCTION 

1.  In  his  study  of  General  Chemistry  the  student  has 
become  somewhat  familiar  with  the  more  important  ele- 
ments, their  preparation,  physical  properties  and  some  of 
their   principal   compounds.     In   his   laboratory   work   he 
has  become  acquainted  with  the  different  types  of  chemical 
reactions;     viz.,     combination,     decomposition,     displace- 
ment,    double    decomposition,     oxidation-reduction,     etc. 
He  has  learned  that  certain  reactions  are  reversible,  that 
others  are  non-reversible,  and  that  the  point  of  equilibrium 
in  reversible  reactions  is  influenced  by  the  relative  amounts 
of  the  substances  entering  into  the  reaction. 

2.  "  Qualitative  Analysis    treats  of  the   identification 
of  matter,"  while  Quantitative  Analysis,  as  its  name  implies, 
deals   with   its   quantitative   determination.     An   efficient 
system    of   qualitative   analysis   should    consist   not   only 
in  finding  out  what  substances  are  present  and  what  are 
absent  but  also  in  obtaining  an  estimate  of  the  relative 
quantity  of  each  constituent  present.     In  order  to  accom- 
plish this  with  accuracy  and  certainty  the  analyst  should 
know  the  principle  involved  in  the  chemical  actions  con- 
cerned,  the  reason  for  each   reagent   used,   the  result  it 
produces  and  how  this  result  is  brought  about.     For  the 
purpose   of   Qualitative   Analysis   it   is    usually   desirable 

1 


2  INTRODUCTION 

and  often  necessary  to  separate  the  substances  from  each 
other  and  then  apply  some  characteristic  chemical  reaction 
which  can  be  easily  and  accurately  recognized. 

3.  In  the  course  of  this  outline  the  following  terms 
will  be  used  frequently,  and  the  student  should  become 
familiar  with  them  and  their  meaning  at  the  outset. 

(a)  A   Reagent    is    a    substance    which    produces    a 
known  reaction  and  is  used  for  the  purpose  of 
obtaining  a  desired  result. 

(b)  A  solution  is  a  homogeneous  mixture,  the  prop- 
erties of  which  change  gradually  with  change  in 
composition.      Aqueous    solutions    are    of    most 
importance  and  are  used  almost  exclusively  in 
this   outline.     The   student   should   learn,    early 
in   his   course,   to  distinguish   between   solutions 
and  colloidal  suspensions. 

(c)  Precipitate  is  a  term  usually  applied  to  a  solid 
substance  which  separates  from  solution  on  the 
addition  of  a  reagent.     It  may  be  finely  divided 
so  that  it  settles  very  slowly  and  is  not  held  by 
a    filter    paper.     Such    a    precipitate  is   termed 
"colloidal."     Precipitates    are    distinguished    as 
crystalline,  flocculent,  curdy,  colloidal,  etc. 

(d)  Residue  is  a  term  usually  applied  to  that  portion 
of  a  solid  left  undissolved  by  a  given  reagent 
or  solvent. 

(e)  Filtration  consists  in  the  mechanical  separation 
of  a  solid  from  a  liquid,  by  means  of  a  suitable 
screen,    which    allows    only    the    liquid    to    pass 
through.     The  liquid  passing  through  is  known 
as  the  filtrate,  while  the  solid  remaining  on  the 
filter  is  called  the  precipitate  or  residue. 

(/)  Decantation  consists  in  carefully  pouring  off  the 
liquid  from  the  solid  which  has  been  allowed  to 
settle.  This  operation  is  employed  with  crystal- 
line precipitates  where  a  thorough  washing  is 


IONIC  THEORY  3 

essential.  The  washing  liquid,  usually  pure 
water  in  small  portions,  is  thoroughly  mixed 
with  the  solid  and  set  aside  until  the  undissolved 
portion  has  settled.  The  clear  supernatant  liquid 
is  then  poured  off.  The  operation  is  repeated 
until  the  solid  is  sufficiently  washed.  In  most 
of  the  operations  of  Qualitative  Analysis  suf- 
ficient washing  may  be  obtained  by  blowing  a 
fine  stream  of  water  from  a  wash  bottle  on  to 
the  filter  containing  the  precipitate  or  residue. 
(g)  Digestion  is  the  term  applied  to  a  reaction 
between  two  or  more  substances  that  are  mixed, 
hot  or  cold,  and  allowed  to  stand  for  some  time 
with  occasional  stirring. 

4.  Ionic  Theory. — In  his  study  of  the  effect  of  the 
solute  on  the  properties  of  solutions,  Raoult  showed  that 
if  a  non-volatile  substance  is  dissolved  in  pure  water  cer- 
tain properties,  viz.,  boiling-point,  freezing-point,  osmotic 
pressure  and  vapor  pressure,  vary  with  the  concentra- 
tion of  the  solute.  Furthermore,  he  showed  that  equi- 
molecular  proportions  of  certain  substances  produce  an 
equal  effect  on  these  properties,  e.g.,  one  molecular  weight 
of  cane  sugar  (Ci2H22Oii  =342  gms.)  or  of  glucose 
(C6Hi2O6  =  i7i  gms.)  or  of  glycerine  (C3H5(OH)3  =  92 
gms.),  when  dissolved  in  10  liters  of  water,  lowers  the 
freezing-point  0.186°  C.  and  raises  the  boiling-point 
0.052°  C.  From  Avogadro's  Law  we  learn  that  equal 
volumes  of  gases  under  the  same  conditions  of  temperature 
and  pressure  contain  an  equal  number  of  molecules.  A 
careful  study  of  solutions  has  shown  that  a  dissolved 
substance  possesses  properties  similar  to  those  of  a  gas 
having  the  same  molecular  concentration.  A  consider- 
ation of  these  and  other  facts  leads  to  the  conclusion  that 
equimolecular  proportions  of  all  substances,  whether  liquid, 
solid  or  gas,  contain  equal  numbers  of  molecules.  The 
change  in  freezing-point,  boiling-point,  etc.,  is  therefore 


4  INTRODUCTION 

proportional  to  the  number  of  molecules  or  particles  of 
solute  dissolved  in  a  given  amount  of  the  solvent.  Examina- 
tion of  a  great  variety  of  substances  in  different  solvents 
has  demonstrated  the  truth  of  this  conclusion. 

5.  When  an  acid,  base  or  salt  is  dissolved  in  pure  water, 
the  change  in  freezing-point,  boiling-point,  etc.,  is  greater 
than  would  be  expected  from  the  general  rule.     In  their 
chemical  relations  these  compounds  show  a  marked  dif- 
ference from  those  which  follow  the  rule.     Their  reactions 
in    solution    are    very    rapid.     In    double    decomposition 
reactions  they  seem  to  be  composed  of  two  or  more  radicals 
which  act  largely  independently  of  each  other.     They  are 
the  only  substances  whose  solutions  conduct  electricity. 
Furthermore,  there  are  certain  colored  salts,  e.g.,  copper 
chloride  (CuCU,  greenish-yellow),  copper  bromide  (CuBr2, 
reddish-brown),  and  copper  sulphate  (CuSC^,  blue),  whose 
solutions  on  being  diluted  finally  assume  the   same  color 
(blue).     In  explanation  of  these  facts  Arrhenius,   in  1885, 
proposed    what    is    known    as    the    "Ionic    Theory.'11     It 
assumes  that  acids,   bases  and   salts,   when  dissolved   in 
water,  dissociate  into  two  or  more  radicals  or  particles, 
that  these  particles  carry  an  electric  charge  and  that  an 
equilibrium  exists  between  the  undissociated  particles  and 
their  dissociation  products. 

6.  When  an  electric  current  is  passed  through  a  solu- 
tion of  an  acid,  e.g.,  HC1,  H2SO4,  etc.,  the  hydrogen  collects 
around  the  negative  electrode  and  the  remaining  radical, 
(Cl),    (SOO,  etc.,    collects  around  the  positive  electrode. 
When  the  current  is  made  to  pass  through  a  solution  of 
a   base,    e.g.,    NaOH,    KOH,   etc.,   the   hydroxide   radical 
(OH)    proceeds   toward   the   positive  electrode   while   the 
metal   radical   is   carried   toward   the   negative   electrode. 
In  the  case  of  salt  solutions,   e.g.,   NaCl,   Na2SO4,   etc., 
the  metal  and  acid  radicals  act  in  the  same  manner  toward 
the  electric  current  as  if  they  were  present  as  the  base  and 
acid  respectively.     A  radical,  therefore,  proceeding  under 
the  influence  of  an  electric  current,  always  moves  toward 


KINETIC  THEORY  AND  IONIC  EQUILIBRIUM  5 

the  same  pole  regardless  of  whether  it  is  present  as  an  acid, 
base  or  salt.  From  this  it  follows  that  the  hydrogen  radical 
must  carry  a  positive  electric  charge  and  the  hydroxide 
radical  a  negative  charge.  Similarly  the  metal  radical 
will  be  charged  positively  and  the  acid  radical  negatively. 
A  radical  bearing  an  electric  charge  is  called  an  ion  and  the 
process  by  which  ions  are  formed  from  the  undissociated 
molecules  is  called  ionization. 

7.  Kinetic  Theory  and  Ionic  Equilibrium.  —  When  an 
acid,  base  or  salt  goes  into  solution  the  influence  of  the 
water  causes  it  to  dissociate,  with  the  formation  of  posi- 
tively and  negatively  charged  ions.  On  the  basis  of  the 
kinetic  theory  we  may  assume  that  the  continual  move- 
ment and  jostling  about  of  the  molecules  causes  them  to 
split  apart  or  dissociate.  At  the  same  time  the  dissociated 
particles  or  ions  will  collide  with  each  other  with  the  result 
that  some  of  them  will  again  unite  to  form  undissociated 
molecules.  At  first  the  former  reaction,  dissociation,  is 
more  rapid,  but  as  the  number  of  ions  increases  their 
union  becomes  more  frequent  until  after  a  time  the  rate 
of  the  two  reactions  will  be  equal,  i.e.,  the  number  of 
molecules  dissociating  in  a  unit  of  time  will  be  just  equal 
to  the  number  formed  by  the  union  of  ions.  When  this 
condition  is  established  the  reaction  is  said  to  be  in  equi- 
librium. It  is  expressed  by  means  of  an  equation  as  follows: 

—  A+B 


8.  Degree  of  Ionization.  —  Experiment  has  shown  that 
the  proportion  of  a  substance  existing  in  the  form  of  ions 
depends  on  its  concentration  in  the  solution,  the  more 
dilute  solution  having  the  higher  per  cent  of  ionization. 
This  is  explained  on  the  basis  of  the  kinetic  theory,  as 
follows:  In  concentrated  solutions  the  ions  are  compar- 
atively close  together  and  collisions  will  be  relatively 
frequent,  while  in  the  more  dilute  solutions  the  ions  are 
necessarily  farther  apart  and  the  time  between  collisions 
will  be  greater,  The  result  is  that  fewer  of  them  will 


6 


INTRODUCTION 


unite,  per  unit  of  time,  to  form  t  non-ionized  molecules. 
The  per  cent  existing  in  the  form  of  ions  must  therefore 
increase  with  increased  dilution,  a  condition  which  agrees 
with  experimental  data.  While  it  has  been  found  that 
the  degree  of  ionization  in  solutions  made  from  salts  is 
relatively  high  and  approximately  the  same  1  for  all  salts 
of  the  same  concentration,  a  wide  variation  exists  in  the 
cases  of  acids  and  bases. 

9.  The  degree  of  ionization  in  normal  and  o.i  normal 
solutions  of  the  more  common  acids,  bases  and  salts  is 
given  in  the  following  table: 


Acids 

PER  CENT  IONIZED 

Bases 

PER  CENT  IONIZED 

N  soln. 

o.i  N    oln. 

N  soln. 

o.i  N  soln. 

H+cr 

H  +  NO3~ 
H+HSO4- 
H+HC2O4~ 

H+C2H302- 
H+HCO-T 

78.4 
82 
5i 

0.41 
0.17 

90 
90 
60 
50(0.2  N) 
(soln.) 
1-3 

K+OH- 
Na+OH- 
Ba  +  +  (OH)2- 
NH4+OH- 

77 
73 
69 
0.4 

86 
86 

I-3I 

SALTS 
Approximate  degree  of  ionization  for  salts  in  o.i  N  solution 

Type  M+A-(e.g.,  KC1) 86  per  cent 

Type  M  +  +A2-(e.g.,  BaCl2) 72  per  cent 

Type  M2+A       (e.g.,  K2SO4) 72  per  cent 

Type  M  +  +A      (e.g.,  BaSO4) 45  per  cent 

10.  Law  of  Mass  Action. — From  a  careful  study  of 
chemical  equilibria  and  rate  of  chemical  action  Gulberg 
and  Waage  showed  that  the  speed  of  a  reaction  is  directly 
proportional  to  the  concentration  of  the  reacting  sub- 
stances. This  is  known  as  the  Law  of  Mass  Action. 

1  Notable  exceptions  to  this  rule  are  HgCl?  and  Pb(C2HgO2)2,  whose  per 
cents  of  ionization  are  relatively  very  small. 


LAW  OF  MASS  ACTION  7 

When  sodium  chloride  (NaCl)   is  dissolved  in  water  the 
following  equilibrium  is  established ; 


Since  the  number  of  NaCl  molecules  which  dissociate  in 
a  unit  of  time  is  proportional  to  the  concentration  of  the 
undissociated  molecules,  the  speed  of  dissociation  may 
therefore  be  stated  mathematically  as  follows: 

Si  =/lCNaCl 

where  Si  =  speed  of  dissociation,  CNaci  =  molar  concentra- 
tion of  undissociated  NaCl  and  f\  =the  proportionality 
constant.  In  like  manner  the  number  of  Na+  and  Cl~ 
which  unite  to  form  the  undissociated  NaCl  per  unit  of 
time  is  proportional  to  the  product  of  their  concentrations 
and  may  be  expressed  mathematically  as  follows: 

S—  f  c*    +  c*  — 
2  —  J  2v^Na     '  v_xci 

where  S2  =  speed  of  union,  CNa+  and  CCi~  =the  con- 
centrations of  the  Na+  and  Cl~  respectively  and  /2  =  the 
proportionality  constant.  When  equilibrium  is  established 
the  speeds  of  the  opposing  actions  must  be  equal ;  therefore 

/iCMaci  =/2CNa+  '  Ccf~ 

and  by  transposition 

CNa+-CCi-     fi 

CNaCl  /2 

Since  /i  and  /2  are  constants  their  ratio  must  be  constant 

and 

C+  c*  - 
Na     *  V^Ci  -IT- 

— P =  K 

^NaCl 

where  K  is  a  constant.  This  is  the  mathematical  state- 
ment of  the  Law  of  Mass  Action  as  applied  to  the  ioniza- 
tion  of  sodium  chloride,  and  shows  that  the  product  of  the 
concentrations  of  the  ions  in  the  solution  divided  by  the 
concentration  of  the  non-ionized  molecules  is  a  constant 


8  INTRODUCTION 

quantity  which  is  independent  of  the  source  of  the  ions. 
K  is  known  as  the  ionization  constant.  For  highly 
ionized  substances  that  are  very  soluble  K  varies  consider- 
ably with  the  change  in  concentration,  but  for  slightly 
ionized  substances  and  those  that  are  difficultly  soluble 
it  remains  practically  the  same.  The  following  examples 
will  serve  to  illustrate. 

Experiment  has  shown  that  in  a  molar  solution  of 
acetic  acid  (HC2H3O2),  0.41  per  cent  of  the  acid  is  in  the 
form  of  ions  while  99.59  per  cent  of  it  remains  in  the 
non-ionized  state.  The  ion  concentration,  therefore,  is 
I  X  0.004  1  =0.0041  and  that  of  the  non-ionized  portion 
is  1X0.9959=0.9959,  i.e.,  the  condition  of  equilibrium 
in  a  liter  of  i  molar  acetic  acid  becomes 

HC2H302    -         H+          +     C2H302- 
(0.9959  m°l)      (0.0041  mol)     (0.0041  mol) 

Substituting  these  values  in  the  Mass  Law  equation 
r*  +  r* 

^-H      '  V- 


K  is  found  to  have  the  value 

0.0041  Xo.oo4i=OQ 
0-9959 

If  the  above  molar  solution  is  diluted  to  ten  times  its 
original  volume  the  proportion  of  acid  that  exists  in  the 
form  of  ions  will  increase  to  1.3  per  cent.  Using  this 
value  we  obtain  0.1X0.013=0.0013  for  the  ion  concen- 
tration and  0.1X0.987=0.0987  for  the  concentration  of 
the  non-ionized  portion.  Substituting  these  values  in  the 
Mass  Law  equation 

CH+  •  CC2H302-  =  0.0013  xo.ooi3       ooQQ 

CHC2H302  0.0987 

a  value  is  obtained  which  is  in  very  good  agreement  with 
that  obtained  in  the  molar  solution. 

In  the  case  of  ammonium  hydroxide  (NH4OH)  experi- 


SOLUBILITY  PRODUCT  9 

ment  shows  that  a  o.i  molar  solution  is  1.31  per  cent 
ionized.  The  ion  concentration,  therefore,  is  o.i  X 0.0131  = 
0.00131  and  the  concentration  of  the  non-ionized  molecules 
is  o.i  Xo. 9869  =0.09869.  Substituting  these  values  in  the 
mass-law  equation  we  obtain  for  the  ionization  constant 

CW.CaH-  =o.ooi3i  XQ.OOI3I  =0.OOOOI73 

CNH4OH  0.09869 

In  a  o.oi  molar  solution  the  per  cent  of  ionization  has  been 
found  to  be  4.07,  from  which  we  obtain  the  following 
equation 

(o  .01X0. 0407)  X  (o .  01  Xo .  0407)     T^ 

— T — — r ^—^  =K  =0.00001 72 

(o.oi  Xo.9593) 

ii.  Solubility  Product. — Since  precipitation  methods 
play  an  important  part  in  Analytical  Chemistry,  a  con- 
sideration of  the  mass  law  in  its  relation  to  saturated 
solutions  is  of  considerable  importance.  When  a  sub- 
stance, such  as  sugar  or  salt,  is  placed  in  contact  with  a 
liquid,  some  of  the  molecules  of  the  solid  enter  the  liquid 
and  in  accordance  with  the  kinetic  theory  move  about 
in  all  directions  within  the  liquid.  After  a  time  some  of 
them  will  return  to  the  solid,  and  as  more  of  the  solid 
dissolves  the  number  of  molecules  returning  to  the  solid 
will  increase,  until  the  number  entering  the  liquid  and 
the  number  leaving  it  in  a  unit  of  time  are  equal.  When 
this  condition  prevails  equilibrium  is  established  and  the 
number  of  dissolved  molecules  (molar  concentration)  is 
a  constant.  The  solution  is  said  to  be  saturated.  It 
follows,  therefore,  that  in  a  saturated  solution  of  an  acid, 
base  or  salt  there  must  be  an  equilibrium  between  the 
undissolved  solute,  the  undissociated  molecules  in  solu- 
tion and  its  ions.  In  the  case  of  the  difficultly  soluble 
salt,  AgCl,  this  equilibrium  may  be  expressed  as  follows: 

AgCl1-AgCl-Ag++Cl- 

1 A  line  ( )  drawn  beneath  a  symbol  will  be  used  to  denote  the 

undissolved  solid,  precipitate  or  residue. 


10  INTRODUCTION 

and  the  equilibrium  for  the  Law  of  Mass  Action  becomes 

CAg+  •  CCi~  =j^ 

CAgd 

Since  the  solution  is  saturated,  CAgCi  is  a  constant  quantity 
and 

CAg+-Ccr  =K-CAgCi=K 

Therefore,  in  a  saturated  solution  of  a  given  ionogen  the 
product  of  the  concentration  of  its  ions  1  is  a  constant  and 
is  called  the  Solubility  Product. 

In  the  following  table  are  placed  the  solubility  products 
at  1 8°  C.  of  some  of  the  more  common  substances  met 
with  in  qualitative  analysis. 


Substance 

K 

Substance 

K 

HgS 

4.o-io-53 

Ag2Cr04 

i.o-io-12 

CuS 

8.5-IO-45 

Mg(OH)2 

3.4-IO-11 

CdS 

3.6-IO-29 

BaCrO4 

I.6-IO-10 

PbS 

4.2-IO-28 

BaSO4 

9.0-  io~10 

CoS 

3.0-IQ-26 

AgCl 

8.7-IO-9 

NiS 

I.4-IQ-24 

CaC2O4 

i.y-io-9 

ZnS 

I.2-IO-23 

CaCO3 

2.8-IO-9 

FeS 

i.5-io-19 

BaCO3 

I.9-IQ-9 

MnS 

I.4-IQ-15 

PbSO4 

i.o-io-8 

MgNH4PO4 

2.5-IO-13 

SrSO4 

2.8-IO-7 

The   ion   product,    and   therefore   the   solubility,   of  a 
substance  may  be  altered  in  the  following  ways: 

(a)  By    the    addition    of    a    reagent    containing   a 
common  ion. 

(b)  By  addition  of  a  reagent  which  forms  with  one 
of  the  ions  a  slightly  ionized  compound. 

1  This  is  true  in  the  case  of  ionogens  consisting  of  one  cation  and  one  anion. 
In  other  ases  the  solubility  product  should  contain  the  ion  concentration 
raised  to  a  power  equal  to  the  number  of  ions  that  are  alike  in  its  formula, 
e.g.,  the  solubility  product  for  PbCl2  should  be  written  Cpb+  +  -C2ci~  =K. 


COMMON   ION  11 

(c)  By  addition  of  a  reagent  which  unites  with  one 
of  the  ions  to  form  a  complex  ion. 

(d)  By  addition  of  a  reagent  which  alters  the  charge 
on  one  on  the  ions.     (Oxidation  and  reduction.) 

(e)  By  addition  of  a  strong  acid  or  strong  base  to 
an  amphoteric  substance. 

12.  Common  Ion.  —  When  a  soluble  chloride,  such  as 
HC1  or  NH4C1,  is  added  to  an  acid  solution  of  a  silver  salt, 
silver  chloride  (AgCl)  is  precipitated.  The  solution  will 
be  saturated  with  respect  to  silver  chloride  when  the  ion 
product,  CAg+-CCi~,  reaches  the  solubility  product  value, 
KAgci,  for  silver  chloride.  Any  further  addition  of  chloride 
tends  to  increase  the  ion  product  above  the  solubility 
product  value.  This  in  turn  disturbs  the  equilibrium 

AgCl-Ag++Cl- 

so  as  to  decrease  the  concentration  of  silver  ions.  Since 
the  concentration  of  the  silver  ion  can  be  decreased  only 
by  uniting  with  chloride  ions  to  form  non-ionized  AgCl 
the  solution  tends  to  become  supersaturated  with  respect 
to  AgCl  molecules.  Some  of  the  silver  chloride  will  there- 
fore be  precipitated  and  the  following  equilibrium  will 
result  : 


At  25°  the  ion  concentration  of  a  saturated  solution  of 
silver  chloride  has  been  found  to  be  1.2  -io-5.  From  this 
is  obtained  for  the  solubility  product, 


CAg+  •  Cci~  =  K  =  i  .44  •  io- 


10 


If  to  I  liter  of  this  solution  o.oi  mole  (0.535  gmO  of 
ammonium  chloride  (NH4C1)  (86  per  cent  ionized)  is  added, 
the  concentration  of  the  chloride  ions  is  increased  by 
0.01X0.86=0.0086,  and  Ccr  will  be  0.0086+0.000012  = 
0.008612  or  about  700  times  as  large  as  in  the  original 
solution.  Since  CAg+'Ccr  =  144- io~10  it  will  be  seen 


12  INTRODUCTION 

that  the  concentration  of  silver  ions  must  decrease  to  about 
TU~O  of  its  original  value.  The  student  should  especially 
note  that  the  completeness  with  which  a  given  ion  may 
be  removed  from  solution  in  this  way  depends  on  the 
concentration  of  the  non-ionized  molecules  in  a  saturated 
solution,  since  their  concentration  is  not  decreased  by 
addition  of  a  common  ion.  Complete  precipitation,  from 
the  standpoint  of  Analytical  Chemistry,  is  obtained  only 
when  this  value  is  very  small. 

13.  Slightly  Ionized  Compounds.  —  The  preceding  para- 
graph has  shown  how  the  addition  of  a  common  ion  may 
be  used  to  remove  a  given  ion  from  solution  when  the 
compound  formed  is  difficultly  soluble.  In  the  same  way 
a  given  ion  may  be  reduced  to  almost  nothing  when  the 
compound  formed  is  soluble  but  very  slightly  ionized. 
If  to  a  o.i  molar  solution  of  acetic  acid  (HC2H3O2)  (1.3 
per  cent  ionized)  a  o.i  mole  of  some  soluble  acetate,  such 
as  sodium  acetate  (NaC2H3O2)  (86  per  cent  ionized),  is 
added,  the  large  excess  of  acetate  ions  tends  to  increase 
the  speed  of  union  of  H+  and  C2H3O2-  and  so  shifts  the 
equilibrium  HC2H3O2^±H+  +C2H3O2-  to  the  left.  From 
the  equation 

o2-  =o.  0013X0.  0013  =  I   7.IO-5 


CHC2H3o2  0.0987 

it  will  be  seen  that  CHC2H302  cannot  be  appreciably  increased 
owing  to  the  small  concentration  of  H+  available.  The 
product  CH+'CC2H3o2~  must  therefore  recover  approxi- 
mately its  original  value.  Since  on  the  addition  of  sodium 
acetate,  CC2H3o2~  becomes  0.086+0.0013  =  0.0873  or  about 
60  times  its  original  value,  CH+  must  be  decreased  to 
about  ^o  of  its  former  magnitude.  While  the  per  cent 
of  ionization  of  the  salt  must  also  decrease  because  of  the 
presence  of  the  acetate  ions  from  the  acid,  the  amount  is 
negligible  in  proportion  to  its  original  value  since  the 
number  of  acetate  ions  is  relatively  so  few  (0.0013  :  0.086). 
Therefore  when  the  compound  formed  is  soluble  a  given 


HYDROLYSIS  13 

ion  can  be  reduced  to  almost  nothing  only  when  the  com- 
pound is  very  slightly  ionized. 

14.  Hydrolysis. — Pure  water  ionizes  to  a  slight  extent 
into  Hf  and  OH~.  Although  the  ions  of  water  may  be 
neglected  when  all  the  substances  concerned  in  a  given 
reaction  are  highly  ionized,  they  become  quite  appreci- 
able and  must  be  taken  into  consideration  when  the  re- 
action involves  substances  that  are  very  slightly  ionized 
or  difficultly  soluble.  Experiment  has  shown  that  the 
concentration  of  H+  and  therefore  of  OH+  in  pure  water 
at  25°  is  io~7.  The  concentration  of  the  non-ionized 
molecules  is  therefore  very  large  in  comparison  and  may 
be  considered  constant.  From  the  Law  of  Mass  Action, 
then,  the  product  of  the  concentrations  of  the  ions  becomes 
a  constant.  Since  CH+ =COH~  =IO"7  we  have  for  the 
ion  product  CH+  -C0H~  =  io-14.  In  any  solution,  there- 
fore, the  concentration  of  H+  multiplied  by  the  concen- 
tration of  OH~  must  equal  the  ion  product  constant, 
io~14.  An  increase  of  H+  must  result  in  a  decrease  of 
OH-  and  vice  versa. 

When  sodium  acetate  (NaC2H3O2),  a  highly  ionized 
salt,  is  dissolved  in  water  the  concentration  of  C2HaO2~ 
may  become  so  large  that  its  product  with  the  H+  of  the 
water  will  exceed  the  ionization  value  for  the  slightly 
ionized  acetic  acid  (H^HsC^).  Some  of  the  C2HaO2~ 
and  H+  therefore  unite  to  form  the  non-ionized  acid 
which  momentarily  reduces  the  ion  product  for  water 
below  io~14.  The  result  is  that  more  water  ionizes  until 
the  product  CH+-COH~  again  reaches  io~14.  The  student 
should  note  that  CH+  is  now  less  than  COH~  and  therefore 
the  solution  becomes  basic.  On  the  other  hand,  if  ferric 
chloride  (FeCls)  is  dissolved  in  water  the  concentration 
of  Fe+++  from  the  highly  ionized  salt,  multiplied  by  the 
concentration  of  OH~  already  present  in  the  water,  may 
exceed  the  ionization  value  for  the  very  slightly  ionized 
ferric  hydroxide  and  form  non-ionized  Fe(OH)3.  The 
ion  product  for  water  is  thus  momentarily  reduced  below 


14  INTRODUCTION 

io~14.  More  water  must  therefore  ionize  until  the  prod- 
uct CH+  -C0H~  =  io~14  is  reached.'  CH+  is  now  greater 
than  COH~  and  the  solution  reacts  acid.  It  will  be  seen 
therefore  that  the  ions  of  water  must  be  taken  into  con- 
sideration when  a  substance  is  involved  either  one  of 
whose  ions  may  unite  with  one  of  the  ions  of  water  to  form 
a  very  slightly  ionized  compound.  Should  the  compound 
be  difficultly  soluble  the  equilibrium  may  be  shifted  to 
completion  and  a  given  ion  removed  from  solution. 

15.  Complex  Ion.  —  When  ammonium  hydroxide  is 
added  to  a  solution  of  a  copper  salt,  copper  hydroxide 
(Cu(OH)2)  is  at  first  precipitated,  but  on  the  addition  of 
an  excess  of  the  reagent  the  Cu(OH)2  precipitate  is  dis- 
solved and  a  deep  blue  solution  is  obtained.  It  would 
seem  from  the  foregiong  discussion  and  the  principle  of 
the  Law  of  Mass  Action  that  an  excess  of  the  reagent 
should  produce  a  more  complete  precipitation  as  was  found 
to  be  the  case  with  silver  chloride  (see  12  above).  It  will 
be  remembered  from  the  study  of  general  Chemistry  that 
when  ammonia  (NH3)  is  dissolved  in  water  only  a  small 
portion  of  it  reacts  with  the  water  to  form  ammonium 
hydroxide  (NH4OH),  the  greater  part  of  it  remaining  in 
the  solution  as  ammonia  (NHs).  The  following  equilib- 
rium must  therefore  exist  in  the  solution: 


H2O  +NH3^  NH4OH  —  NH4+  +OH-. 

An  examination  of  the  deep-blue  copper  solution  .shows 
the  presence  of  the  complex  ion  Cu(NH3)4++.  From  the 
principle  of  the  solubility  product 


it  is  evident  that  any  increase  of  OH~  above  that  necessary 
to  reach  the  solubility  product  for  Cu(OH)2  must  result 
in  a  decrease  in  the  Cu++  concentration.  The  concen- 
tration of  the  Cu++  may  be  decreased  either  by  the  form- 
ation of  non-ionized  Cu(OH)2  and  consequent  precipitation 
or  by  its  union  with  free  ammonia  to  form  the  complex 


AMPHOTERIC  SUBSTANCES  15 

ion  Cu(NH3)4++.  The  high  proportion  of  free  ammonia 
and  the  slight  dissociation  of  the  complex  ion  Cu(NH3)4++ 
both  influence  the  equilibrium  toward  the  formation  of 
the  complex  ion,  hence  the  net  result  is  that  the  equilibrium 

Cu(OH)2-  Cu(OH)2-  Cu++  +20H- 


Cu(NH3)4 


++ 


will  shift  toward  the  formation  of  the  complex  ion  and  the 
ion  product  will  be  decreased  below  that  of  the  solubility 
product  value;  more  Cu(OH)2  will  dissociate  and  the 
precipitate  will  pass  into  solution.  The  use,  therefore, 
of  a  reagent  which  will  react  with  a  given  ion  to  form  a 
complex  ion  may  be  made  in  order  to  bring  a  substance 
into  solution,  to  prevent  precipitation  or  to  remove  an 
ion  from  the  field  of  action. 

16.  Amphoteric  Substances.  —  An  amphoteric  element 
is  one  whose  hydroxide  in  solution  ionizes  both  as  an  acid 
and  as  a  base,  i.e.,  it  produces  both  hydrogen  and  hydroxyl 
ions.  When  a  strong  acid,  such  as  HC1,  is  added  to  a 
precipitate  of  aluminium  hydroxide  (A1O3H3)  the  pre- 
cipitate is  dissolved  and  experiment  shows  that  the  alu- 
minium is  present  in  the  solution  as  the  positive  aluminium 
ion  (Al+++).  On  the  other  hand,  when  a  strong  base, 
such  as  NaOH,  is  added  to  the  aluminium  hydroxide 
precipitate,  the  precipitate  is  dissolved;  but  experiment 
shows  that  the  aluminium  is  present  in  the  solution  as 
negative  aluminate  ions  (A1O2~).  The  following  equi- 
librium is  therefore  assumed  to  exist  in  a  neutral  solution 
of  aluminium  hydroxide  : 

A1+++  +3OH-  —  A1O3H3  ^  H+ 

+H2A1O3-  ^  H+  +A1O2-  +H2O 

When  a  strong  acid,  furnishing  its  high  concentration  of 
H+,  is  added  the  above  equilibrium  is  disturbed  owing 


16  INTRODUCTION 

to  the  union  of  H+  with  the  OH~  present  to  form  the 
very  slightly  ionized  water.  Non-ionized  A1O3H3  then 
dissociates  further  to  produce  more  OH~  with  the  final 
result  that  the  aluminium  hydroxide  is  dissolved  and  the 
aluminium  remains  in  solution  as  A1+++;  i.e.,  the  equi- 
librium shifts  to  the  left  and  A1O3H3  acts  as  a  base.  When 
a  strong  base  is  added  the  high  concentration  of  OH~ 
tends  to  use  up  the  H+  present  in  forming  water  as  above. 
This  causes  a  further  dissociation  of  A1O3H3  to  produce 
more  H+,  with  the  final  result  that  the  aluminium  hydrox- 
ide is  dissolved  and  the  aluminium  remains  in  the  solu- 
tion as  A1O2~;  i.e.,  the  equilibrium  shifts  to  the  right 
and  A1O3H3  acts  as  an  acid.  It  should  be  noted  that  if 
a  -weak  base,  such  as  NH4OH,  is  substituted  for  the  strong 
base  mentioned  above  the  effect  will  be  very  much  less 
noticeable,  owing  to  the  much  smaller  concentration 
of  OH-. 

17.  Oxidation  and  Reduction.  —  When  iron  is  acted 
upon  by  hydrochloric  acid,  hydrogen  is  displaced  and  iron 
passes  into  solution  according  to  the  following  equation: 

Fe+2HCl-»FeCl2+H2 

Considered  from  the  ionic  standpoint  this  gives 
Fe  +2H+  +2Cl-->  Fe++  +2C1-  +H2 


Now  if  a  stream  of  chlorine  gas  is  passed  through  the  solu- 
tion a  further  change  takes  place  as  follows: 

C12  +2Fe++  +4Cl--»  2Fe+++  +6CK 


In  passing  into  solution  the  iron  has  become  positively 
charged,  while  at  the  same  time  charged  hydrogen  has 
become  neutral  and  neutral  chlorine  has  become  negatively 
charged. 

A  free  atom  may  be  said  to  consist  of  a  positively 
charged  nucleus  surrounded  by  a  number  of  negatively 


OXIDATION  AND  REDUCTION  17 

charged  particles  called  electrons.  These  electrons  are 
capable  of  existing  independently  of  the  atom  and  hence 
may  leave  one  atom  and  attach  themselves  to  another. 
The  mechanism,  therefore,  of  the  above  reactions  may 
be  stated  briefly  as  follows:  An  atom  of  iron,  capable  of 
losing  electrons,  comes  in  contact  with  a  hydrogen  ion 
(hydrogen  atom — one  electron).  Two  electrons  leave  the 
atom  of  iron  and  attach  themselves  to  two  hydrogen  ions, 
which  are  in  turn  neutralized.  The  loss  of  these  two 
electrons  has,  therefore,  left  the  iron  positively  charged. 
In  the  second  reaction  a  chlorine  atom,  capable  of  holding 
an  additional  electron,  comes  in  contact  with  an  iron  ion 
and  receives  an  electron  from  it.  The  chlorine,  therefore, 
becomes  negatively  charged  while  the  iron  remains  with 
a  higher  positive  charge. 

The  quantity  of  electricity  equivalent  to  that  carried 
by  an  electron  is  called  a  "  unit  charge  "  and  may  be 
either  positive  or  negative  in  character.  The  number 
of  excess  "  unit  charges  "  carried  by  an  atom  or  ion  is 
numerically  equal  to  its  valence.  Valence,  therefore, 
may  be  either  a  positive  or  a  negative  number  depending 
on  whether  the  atom  or  ion  holds  less  or  more  electrons 
than  is  sufficient  to  neutralize  the  positive  nucleus.  It 
follows,  therefore,  that  the  valence  of  an  element  in  the 
free  state  is  zero  and  that  the  algebraic  sum  of  the  positive 
and  negative  valences  in  any  compound  is  zero.  Oxida- 
tion consists  in  the  loss  of  one  or  more  electrons  by  an 
atom  or  ion,  i.e.,  an  algebraic  increase  in  valence.  Reduc- 
tion consists  in  the  addition  of  electrons  to  an  atom 
or  ion,  i.e.,  an  algebraic  decrease  in  valence.  It  will  be 
seen,  therefore,  that  oxidation  and  reduction  must  accom- 
pany each  other  and  be  equivalent  in  amount;  i.e.,  in 
a  given  reaction  if  an  element  or  ion  loses  one  or  more 
electrons  those  electrons  must  attach  themselves  to  some 
other  element  or  ion. 

In  writing  equations  of  oxidation  and  reduction  the 
student  should  first  write  the  skeleton  equation;  e.g., 


18  INTRODUCTION 

when  H2S  is  passed  into  a  solution  of  HNOs  free  sulphur 
is  produced  and  the  HNOs  is  reduced  to  nitric  oxide  (NO). 

H2S+HN03->S+NO+H20 

He  should  then  note  any  change  in  valence,  i.e.,  what 
elements  have  lost  or  gained  electrons  during  the  reaction, 
and  the  number  lost  or  gained  by  each  atom.  In  the 
above  equation  sulphur  has  changed  from  a  valence  of 
negative  2  in  H2S  to  zero  in  free  sulphur;  i.e.,  each  atom 
of  sulphur  has  lost  2  electrons.  Nitrogen,  on  the  other 
hand,  has  changed  from  a  positive  valence  of  5  in  HNOs 
to  positive  2  in  NO;  i.e.,  3  electrons  have  attached  them- 
selves to  each  nitrogen  atom.  Since  the  total  number 
of  electrons  lost  by  one  element  must  be  equal  to  the  total 
number  gained  by  the  other,  it  is  evident  that  3  molecules 
of  H2S  will  furnish  just  enough  electrons  to  supply  those 
necessary  to  change  the  nitrogen  in  2  molecules  of  HNOs 
to  NO.  The  balanced  equation,  therefore,  becomes 

3H2S  +2HNO3->  38  +2NO  +4H2O 

For  purposes  of  balancing  equations  of  oxidation  and 
reduction  the  valence  of  combined  hydrogen  should  always 
be  considered  as  positive  i  and  that  of  combined  oxygen 
as  negative  2. 

A  few  of  the  more  important  oxidizing  and  reducing 
agents  are  given  in  the  following  table: 

Oxidizing  Agents  Reducing  Agents 

1.  Halogens  (Cl,  Br,  I)  i.  SnCl2 

2.  HNO3  2.  H2S 

3.  Aqua  regia  3.  Nascent  hydrogen 

4.  KC1O4  4.  SO2 

5-  Na2O2  5.  H2C2O4 

6.  K2Cr2O7  6.  Alcohol 

7.  KMnO4 

8.  Pb02 


LABORATORY  SUGGESTIONS  19 

LABORATORY  SUGGESTIONS 

Qualitative  Analysis  has  to  do  with  both  dry  and  wet 
reactions.  The  dry  reactions  are  those  used  largely  in 
blow-pipe  analysis  and  will  not  be  considered  here.  Since 
wet  reactions  will  be  used  almost  exclusively  in  this  out- 
line a  brief  consideration  of  some  of  the  most  important 
processes  is  given. 

Filtration. — Since  a  finely  divided  precipitate  not  only 
tends  to  pass  through  the  filter  and  so  make  its  separation 
difficult  but  also  tends  to  clog  the  filter  paper  and  thus  render 
the  process  of  filtration  slow  and  tedious,  it  is  necessary  to 
have  the  particles  as  large  as  possible.  This  is  accomplished 
most  effectively  by  adding  the  precipitating  agent  slowly 
to  a  hot  solution.  Whenever  permissible,  time  will  usually 
be  saved  by  filtering  a  solution  while  hot,  since  hot  water 
passes  through  the  filter  paper  more  rapidly  than  cold  water. 

The  rate  of  filtration  is  also  influenced  very  largely 
by  the  position  of  the  filter  paper  in  the  funnel.  The  paper 
should  fit  closely  against  the  funnel  so  that  no  air  passages 
exist  between.  This  may  be  accomplished  by  folding  the 
circular  filter  paper  in  half  and  then  folding  again  as  shown 
in  Fig  I.  The  second  fold  should  be  pinched  together 
at  the  point,  opened  between  the  longer  fold  so  as  to  form 
an  inverted  cone  and  pressed  gently  into  the  funnel  until 
it  fits  snugly  against  the  glass  all  the  way  around.  With- 
out removing  it  from  the  funnel  the  second  fold  may  now 
be  creased  and  the  paper  wet  with  a  stream  of  water 
blown  from  the  wash  bottle  to  hold  it  in  position  (Fig.  2). 

Wash  Bottle. — After  checking  in  his  apparatus  each 
student  should  make  a  wash  bottle,  using  a  500  cc.  or 
750  cc.  Florence  flask.  The  wash  bottle  should  be  made 
sufficiently  compact  so  that  it  can  be  easily  held  and  the 
nozzle  manipulated  with  one  hand.  A  convenient  form 
is  shown  in  Fig.  3. 

Record  of  Results. — The  student  should  keep  a  careful 
and  accurate  record  of  all  results  obtained  in  his  analysis 


INTRODUCTION 


of  "  unknowns."  The  keeping  of  this  record  not  only 
enables  the  student  to  understand  the  principles  involved 
and  to  follow  the  procedures  more  easily,  but  also  it  often 
enables  the  instructor  to  determine  the  causes  of  errors 
and  so  help  the  student  to  avoid  repeating  them.  The 


FIG.  i. 


FIG.  3. 

following  form  for  "  unknowns  "   has  been  found   to  be 
very  easily  kept  and  quite  satisfactory: 


No. 

Substance 

Reagent 

Result 

Conclusion 

i 

Unknown 

NH4C1 

White  ppt. 

Group  I  present 

2, 

Ppt.  i 

2N.  HC1 

White  res. 

PbCl2,  Hg2Cl2,  AgCl 

3 

Fil.  2 

H2S 

Soln. 

No  Bi  or  Sb 

4 

Res.  2 

Hot  H2O 

White  res. 

Hg2Cl2,agCl 

5 

Fil.  4 

H2SO4 

Soln. 

No  Pb 

6 

Res.  4 

NH4OH 

Black  res. 

Hg  present 

7 

Fil.  6 

HNO3 

White  ppt. 

Ag  present 

LABORATORY  SUGGESTIONS  21 

The  student  will  also  find  it  of  considerable  benefit 
to  keep  a  record  of  his  results  obtained  in  preliminary 
experiments  by  underlining  and  noting  the  color  of  all 
precipitates  in  outline  form  similar  to  that  of  Table  I  and 
the  following  tables. 


PART  II 


THE  SYSTEMATIC  ANALYSIS 

Cations — (metal  ions) 
PREPARATION  OF  SOLUTION 

If  the  unknown  substance  is  a  liquid  or  solution 
treat  by  (i);  if  a  solid  non-alloy  treat  by  (2)  and  if  an 
alloy  treat  by  (3). 

For  the  purpose  of  a  complete  qualitative  analysis  the 
unknown  should  be  divided  into  four  parts  as  follows: 

First  part  for  organic    matter    and    general    infor- 
mation. 

Second   part  for  analysis  of  cations   (metal  ions). 
Third  part  for  analysis  of  anions  (acid  ions). 
Fourth  part  for  special  tests  and  in  case  of  accident. 

(1)  Unknown  Liquid. — Test   with   litmus   for  acidity. 
Evaporate  a  known  volume  to  dryness  in  a  porcelain  dish 
and  note  the  amount  of  residue.     If  organic  matter  may 
be  present  test  the  residue  by  (4).     In  case  organic  matter 
is  known  to  be  absent  treat  an  amount  of  the  solution 
which  contains  about  i  gram  of  solid  by  (10). 

(2)  Unknown  Solid  (non-alloy). — Treat  a  small  portion 
(about    o.i  gm.)    of    the   finely   powdered   substance    for 
organic    matter    by   (4).       If    organic    matter    is    abse.nt 
add  to  another  small  portion  in  a  test-tube  10-15  cc.  of 
water,  and  shake  the  mixture  thoroughly.     If  it  fails  to 
dissolve,  heat  to  boiling.     In  case  the  substance  is  insoluble 
in  water  try  to  dissolve  another  small  portion  in  5  cc.  of 

22 


UNKNOWN  SOLID    (NON-ALLOY)  23 

6N.  HNOs.  From  the  knowledge  gained  by  the  above 
tests  treat  about  i  gram  of  the  solid  according  to  (a),  (b) 
or  (c). 

(a)  If  the  substance  is   soluble  in   water   or   dilute 
HNOz,  dissolve  about   i   gram  of  it  using    as 
little  of  the  6N.  acid  as  possible   (see  Discus- 
sion 3)  and  treat  by  (10). 

(b)  If  the  substance  is  soluble  in  dilute  HCl,  dissolve 
about  i  gram.     If  more  than  5  cc.  of  the  6N. 
HC1   is   used   evaporate   the   solution   to   5   cc. 
(see  Discussion  3)  dilute  with  water  to  a  volume 
of  100  cc.  and  treat  by  (20). 

(c)  If  the  substance  is  insoluble  in  dilute  acid,  add 
to  about   i   gram  of  the  finely  powdered  sub- 
stance in  a  porcelain  dish  6  cc.  of  I2N.  HC1, 
cover   the   dish  with  a  watch   glass   and    heat 
gently.       (See    Discussion    4.)       If    a    residue 
remains,    cool,    add   2    cc.    of    i6N.  HNOs   and 
heat  the  mixture.     In  either  case  finally  evapo- 
rate just  to  dryness,  moisten  the  residue  with 
I2N.  HC1    and    again    evaporate    to    dryness. 
Heat  the  residue  to  120—130°  till  it  is  thoroughly 
dry,  keeping  the  dish  in  motion  over  a  small 
flame.     Loosen   the  residue  with  the  end  of  a 
glass  rod,  add  just  5  cc.  of  6N.  HC1  (see  Dis- 
cussion 3)  and  pulverize  with  the  rod  any  large 
particles.     Cover  the  dish  and  warm  the  mix- 
ture, taking  care  that  none  of  the  acid  evapo- 
rates.    Add  10  cc.  of  water  and  heat  to  boiling. 
Filter  while  hot,  treat  the  filtrate  by  (20)  and 
the  residue  by  (5),    (6)   or   (7).     (See  Discus- 
sion 5.) 

(3)  Unknown  Solid  (Alloy). — To  about  0.5  gram  of  the 
finely  divided  material  in  a  porcelain  dish  add  10  cc.  of 
6N.  HNO3,  cover  with  a  watch  glass  and  warm  gently 
as  long  as  the  action  continues,  adding  small  portions  of 


24  CATIONS— (METAL  IONS) 

l6N.  HNOs  from  time  to  time  if  the  action  is  renewed 
thereby.  Finally  evaporate  just  to  dryness,  add  just 
5  cc.  of  6N.  HNO3  and  15  cc.  of  water.  Heat  to  boiling, 
and  if  a  residue  remains,  filter,  treat  the  filtrate  by  (10) 
and  the  residue  by  (2,  c).  (See  Discussion  6.) 

(4)  Organic  Matter. — To  determine  whether  organic 
matter  is  present  in  an  unknown  a  small  portion  of  the 
solid  is  placed  in  a  hard  glass  test-tube  or  in  a  glass  tube 
closed  at  one  end  and  heated  to  dull  redness.  If  the  sub- 
stance chars  (a  black  color  may  be  due  to  certain  metallic 
oxides)  and  emits  a  burnt  odor,  organic  matter  is  present 
and  should  be  removed  as  follows  (see  Discussion  7). 
Place  about  I  gram  of  the  solid  substance  (more  if  the 
amount  of  organic  matter  is  large)  in  a  porcelain  dish 
and  heat  gently  with  5  cc.  of  cone.  H2SO4  until  it  is  well 
charred.  Cool,  add  slowly  and  with  constant  stirring 
i6N.  HNOs  until  violent  action  ceases.  Warm  gently  for 
a  few  minutes  and  then  heat  more  strongly,  keeping  the 
contents  well  stirred,  until  the  substance  is  thoroughly 
charred.  Repeat  the  process  until  the  mixture  becomes 
light  straw  colored  and  remains  so  when  strongly  heated. 
Treat  by  (a)  or  (b). 

(a)  If  the  substance  has  dissolved  completely,  evapo- 
rate under  a  hood  to   1.5   cc.,   cool  and  pour 
the  contents  into  15  cc.  of  water.     If  there  is 
a  residue,  heat  to  boiling  and  boil  as  long  as 
it    seems    to    be    dissolving.     Filter    off    any 
remaining  residue,  wash  thoroughly  and  treat 
by  (6).     Treat  the  filtrate  by  (10). 

(b)  If   the    substance    has    not    dissolved    completely 
transfer  to  a  platinum  crucible  and  treat   by 
(5),  or  if  the  platinum  crucible  is  not  available, 
evaporate  to  1.5  cc.  as  in  the  preceding  para- 
graph, pour  the  contents  into  15  cc.  of  water, 
heat  to  boiling,  filter,  wash  thoroughly  and  treat 
the  filtrate  by  (10).     Treat  the  residue  by  (7). 


TREATMENT  WITH  Na2CO3  SOLUTION  25 

(5)  Treatment  with  H2F2. — Transfer  the  residue  from 
(2,  c)  or  the  mixture  from  (4,  b)  to  a    platinum  crucible, 
add  enough  cone.  H2SO4  to  make  a  total  volume  of  3  cc. 
(see  Discussion  3) ;   heat  the  mixture  with  a  moving  flame 
until  the  thick  white  fumes  of  H2SO4  appear. 

To  test  for  silicate  or  silica  add  carefully  from  a  loop 
of  a  platinum  wire  5—6  drops  of  pure  cone.  H2F2  and  warm 
the  mixture  over  a  steam  bath.  The  formation  of  gas 
bubbles  shows  the  presence  of  silica  or  silicate.  (See 
Discussion  5.) 

Now  add  2-5  cc.  more  of  the  pure  cone.  H2F2,  cover 
the  crucible  and  digest  the  mixture  on  the  steam  bath  for 
about  fifteen  minutes  unless  solution  takes  place  more 
quickly.  Remove  the  cover  and  evaporate  carefully  until 
the  white  fumes  of  H2SO4  appear.  Treat  by  (a)  or  (b). 

(a)  If  there  is  no  residue  or  precipitate  evaporate 
carefully  to  dryness.     If  there  is  still  no  residue, 
or  only  an  insignificant  one,  the  material  con- 
tained only  silica  or  silicate  and  may  be  dis- 
carded. 

(b)  If  there  is  a  residue  or  precipitate  pour  the  con- 
tents of  the  crucible  into  15  cc.  of  water,  rinsing 
out   the   crucible   with   the   resulting   solution. 
Boil   the  mixture  gently  as  long  as  any  of  the 
residue  seems  to  dissolve.     Filter  and  treat  the 
filtrate  by    (10).     Wash   the   residue  with  iN. 
H2SC>4,  rejecting  the  washings,  and  treat  by  (7). 

(6)  Treatment  with  Na2CO3  Solution. — Mix  the  resi- 
due obtained  in  (2,  c)  or  (4,  a)  with  10  parts  of  solid  Na2COs 
and  20  cc.  of  water  and  boil  the  mixture  for  about  five 
minutes.     Filter  and  reject  the  filtrate  (see  Discussion  8). 
Wash  the  residue  and  dissolve  it  by  adding  6N.  HC1  until 
the  solution  remains  acid;    then  add  just  5  cc.  more  of 
the  acid  and  10  cc.  of  water.     Filter  if  necessary  and  treat 
the  filtrate  by  (20).     If  there  is  a  residue  undissolved  by 
this  treatment  treat  it  by  (7). 


26  CATIONS— (METAL  IONS) 

(7)  Fusion  with  Na2CO3. — Mix  the  residue  from  (2,  c), 
(4,  b)  or  (6)  with  ten  times  its  weight  of  Na2CO3  in  a  plati- 
num or  nickel  crucible  (see  Discussion  9)  and  heat  over  a 
very  hot  flame  until  complete  fusion  takes  place.  If 
necessary  to  secure  a  clear  fusion,  add  0.1-0.3  gram  of 
NaNOs  (see  Discussion  10).  Cool,  place  the  crucible  and 
contents  in  a  dish  and  add  carefully  6N.  HC1  until  the 
solution  is  acid.  Evaporate  to  dryness  and  heat  to  120- 
130°  to  render  the  silica  insoluble.  Add  just  5  cc.  of 
6N.  HC1  (see  Discussion  3)  and  10  cc.  of  water  and  heat 
to  boiling.  Filter  to  remove  silica  and  treat  the  filtrate 
by  (20). 

DISCUSSION 

1.  Difficultly   soluble   solids   are   more   easily   brought 
into  solution  if  they  have  previously  been  reduced  to  a 
fine  powder.     This  is  usually  done  by  grinding  the  solid 
in  a  porcelain  or  agate  mortar.     In  the  case  of  very  hard 
substances,   such  as  certain  minerals  and  rocks,  a  heavy 
porcelain  or  iron  mortar  should  be  used  to  reduce  them 
to  small  particles. 

2.  The  preliminary  tests  with  water  and  dilute  acids 
should  be  carried  out,  since  they  furnish  important  indi- 
cations as  to  the  nature  of  the  constituents  present  and 
often  enable  the  analyst,  especially  the  beginner,  to  obtain 
a  solution  more  quickly  and  easily. 

3.  In  order  that   the   solution   may   have   the   proper 
acid  concentration  for  the  precipitation  of  Group  II  metals, 
just  5  cc.  of  6N.  acid  should  be  present.     If,  therefore, 
the  sample  can  be  dissolved  by  the  use  of  5  cc.  of  6N.  acid, 
a  considerable  saving  of  time  will  be  obtained. 

4.  A   mixture   of   HC1   and    HNO3    (3:1),    known   as 
aqua  regia,  is  a  very  powerful  oxidizing  agent  and  is  often 
used  very  effectively  as  a  solvent.     It  should   be   used, 
however,   only  in  case  dilute  acids  or  cone.    HC1   prove 
ineffective,    since    compounds    comparatively    soluble    in 
these  reagents  may  be  rendered  insoluble  by  its  oxidizing 


DISCUSSION  (i-io)  27 

action;  e.g.,  antimony  and  tin  compounds  may  be  oxidized 
to  insoluble  antimonic  oxide  (Sb2O5)  and  metastannic 
acid  (H2SnO3)n  respectively.  The  action  of  I2N.  HC1  is 
noted,  therefore,  before  the  HNO3  is  added. 

Such  substances  as  MnO2  and  PbO2  are  reduced  and 
dissolved  by  I2N.  HC1.  Hot  I2N.  HC1  slowly  dissolves 
such  oxides  as  Sb2O5,  SnO2,  Fe2O3  and  A12O3.  Upon 
the  addition  of  cone.  HNO3,  gold,  platinum  and  HgS  are 
dissolved.  Silver  compounds  are  changed  by  HC1  to 
AgCl,  somewhat  soluble  in  concentrated  acid  but  left 
almost  completely  in  the  residue  on  the  subsequent  addition 
of  dilute  HC1. 

5.  A  residue  undissolved  by  aqua  regia  may  consist 
wholly  of  silica  or  silicates.     The  H2F2  treatment  decom- 
poses  most  silicates  with   the   formation   of   SiF4,   a  gas 
insoluble    in    cone.    H2SO4,    and    hence   volatilizes.     The 
treatment   is,    therefore,    very   effective   for   decomposing 
rocks,    ores    or    other    substances    which    might    contain 
silica  or  silicates. 

The  residue  undissolved  by  H2SO4  may  contain  the 
sulphates  of  barium,  lead,  strontium,  calcium  and  chro- 
mium. It  may  also  contain  bismuth  as  basic  sulphate 
and  antimony  as  Sb2Os  along  with  undecomposed  AgCl. 

6.  Most  alloys  are  attacked  by  cone.  HNO3,  all  of  the 
elements    present   going   into    solution    except    antimony, 
tin,   carbon  and   silicon.     Antimony,    tin  and  silicon  are 
oxidized   to  Sb2Os,    (H2SnO3)ra   and   H2SiO3   respectively, 
all  of  which  form  white  amorphous  precipitates. 

Certain  alloys,  especially  those  containing  iron  and  alu- 
minium, are  more  readily  attacked  by  HC1.  Treatment 
of  the  residue  with  HC1  and  aqua  regia  not  only  brings 
these  alloys  into  solution  but  tends  to  dissolve  the  oxides 
of  tin  and  antimony  formed  by  HNO3.  The  HNO3 
treatment  is  made  first  in  order  to  remove  any  silver  or 
lead  which  would  be  precipitated  as  chlorides  if  aqua  regia 
were  used. 

7.  Certain  kinds  of  organic   matter,   such  as  sugars, 


28  CATIONS— (METAL  IONS) 

tartaric  acid,  etc.,  prevent  the  precipitation  of  aluminium 
and  chromium  hydroxides.  Large  quantities  of  organic 
matter  of  any  kind  interfere  in  precipitations,  filtrations, 
etc.  Therefore,  if  organic  matter  is  present  it  should  be 
removed  before  beginning  the  systematic  analysis.  Organic 
matter  may  be  removed  by  the  H2SO4  and  HNOs  treat- 
ment as  outlined,  or  by  ignition.  The  latter,  however, 
is  inadvisable  in  a  systematic  analysis  since  such  substances 
as  mercury  and  arsenic  are  volatilized  thereby. 

8.  The  sulphates  of  barium,  lead,  strontium,  calcium 
and  bismuth  are  converted  into  carbonates  by  boiling  with 
Na2COs.     A    second    treatment    is    sometimes    necessary 
to  convert  all  the  barium  into  the  carbonate.     The  car- 
bonates are  readily  dissolved  in  HC1.     Anhydrous  chromic 
sulphate  is  converted  into  the  hydroxide  by  boiling  with 
Na2COs.     The  hydroxide  is  soluble  in  HC1.     AgCl  is  only 
slightly  attacked  by  the  Na2COs  solution. 

9.  Most    substances    are    decomposed    and    rendered 
soluble  by  fusion  with  Na2COs,  the  basic  elements  form- 
ing carbonates  and  the  acidic  elements  forming  sodium 
salts.     In  some  cases,  however,  the  carbonate  is  decom- 
posed with  the  formation  of  the  oxide  or  even  the  metal. 
For  this  reason  care  must  be  exercised  in  the  use  of  a  plat- 
inum crucible,  since  these  metals  readily  alloy  with  the 
platinum.     Substances  which  might  contain  any  of  the 
metals  in  Groups  I  and  II  should  not  be  fused  with  an 
alkali  flux  in  platinum.     Although  a  nickel  crucible  may 
be  used,  the  latter  is  attacked  to  such  an  extent  that  a 
subsequent  test  for  this  element  or  for  the  alkali  metals  is 
rendered  unreliable. 

10.  The  addition  of   NaNO3   to  the   Na2CO3   fusion 
serves  to  oxidize  certain  substances  not  acted  upon  by  the 
Na2COa    alone.     Sulphides    are    oxidized    to    sulphates, 
chromium  compounds  to  chromates  and  manganese  com- 
pounds to  manganates.     If  the  fusion  is  to  be  made  in 
platinum   the   quantity   of   NaNOs   added   should   be   as 
small  as  possible. 


PRELIMINARY  EXPERIMENTS  29 

Group  I 

Prelminary  Experiments 
Bi+  ++,  Sb+++,  Pb++,  Hg+,  Ag+ 

In  connection  with  the  following  experiments  study 
Table  I. 

Experiment  i. — Introduce  into  separate  test-tubes 
5  cc.  portions  of  the  test  solutions  containing  the  above 
ions.  Test  for  acidity  with  litmus  paper  and  if  not  already 
acid  make  distinctly  acid  with  6N.  HNOs.  In  each  case 
dilute  with  water  to  a  volume  of  10  cc.  and  add  NEUCl 
as  long  as  a  precipitate  continues  to  form.  Decant,  or 
filter  if  necessary,  and  treat  the  precipitates  as  directed 
in  Exp.  2.  Write  equations. 

NOTES. — Water,  though  very  slightly  ionized,  reacts  with  salts  of  the 
less  basic  metals  to  form  oxy-compounds  or  even  acids.  This  reaction 
is  called  hydrolysis.  The  ionization  of  water  is  prevented  in  large  meas- 
ure by  the  presence  of  strong  acids  or  strong  bases.  Bi  and  Sb  are  both 
weakly  basic  in  character  and  hence  tend  to  react  with  the  water.  This 
tendency  is  aided  either  by  a  high  concentration  of  their  salts  or  by  a  low 
concentration  of  acid.  The  addition  of  NH4C1  as  directed  above,  there- 
fore, may  or  may  not  give  a  precipitate,  since  the  neutral  chlorides  are 
soluble.  (See  Introduction  14).  If  hydrolysis  occurs,  however,  the 
change  may  be  considered  to  proceed  as  follows: 

BiCls->  Bi(OH)2Cl->  BiOCl 

Experiment  2. — To  each  of  the  above  precipitates 
(Exp.  i)  add  5-10  cc.  of  cold  2N.  HC1  and  mix  thoroughly. 
Note  the  solution  of  BiOCl  and  SbOCl.  (Difference, 
separation  of  Bi  and  Sb.)  Filter  off  the  solution  from 
the  PbCl2  and  saturate  the  nitrate  with  H2S.  Saturate 
the  solutions  containing  Bi+  +  +  and  Sb+  +  +  with  H2S. 
Write  all  equations. 

Experiment  3. — Add  to  the  precipitates  of  PbCl2, 
Hg2Cl2  and  AgCl  (Exp.  2)  10  cc.  of  boiling  water.  Note 
the  solution  of  PbCl2.  (Difference,  separation  of  Pb.) 


30  CATIONS—  (METAL  IONS) 

Experiment  aa.  —  Divide  the  PbCl2  solution  (Exp.  3) 
into  two  parts  and  add  K2Cr2O?  to  the  one  and  H2SO4 
to  the  other.  Note  the  color  and  nature  of  the  precipi- 
tates formed.  Write  equations.  Which  of  the  above 
reagents  should  give  the  more  satisfactory  test  for  Pb++. 

Experiment  4.  —  Decant  the  liquid  from  the  residues 
of  Hg2Cl2  and  AgCl  (Exp.  3)  and  add  NH4OH  as  long 
as  the  reaction  seems  to  continue.  Note  the  solution  of 
AgCl.  (Difference,  separation  of  Hg.) 

NOTES.  —  In  the  presence  of  NH4OH,  Hg2Cl2  is  changed  by  auto- 
oxidation  to  white  HgNH2Cl  and  black,  finely  divided  Hg.  A  part  of 
the  mercury  in  its  reduction  to  the  metallic  state  oxidizes  the  remainder 
to  the  bivalent  condition.  The  compound  HgNH2Cl  may  be  considered 
to  be  formed  from  HgCl2  by  replacing  a  chlorine  atom  by  the  univalent 
radical  (NH2)  .  The  reaction  proceeds  as  follows  : 

Hg2Cl2  +  2NH4OH-+  HgNH2Cl  +Hg  +NH4C1  +  2H2O. 

A  solution  of  NH4OH  contains  a  considerable  proportion  of  NH3 
molecules  which  unite  readily  with  Ag+  to  form  the  complex  ion 
Ag(NH3)2+.  Therefore,  when  AgCl  is  treated  with  NH4OH  the  Ag+ 
is  removed  by  the  formation  of  the  complex  Ag(NH2)2+  which  shifts 
the  equilibrium  toward  the  formation  of  more  Ag+  (Le  Chatelier's  Prin- 
ciple) with  the  result  that  the  AgCl  is  dissolved.  The  reaction  proceeds 
as  follows: 

^  Ag(NH3)2Cl 


Experiment  5.  —  Acidify  the  solution  of  Ag(NH3)2Cl 
(Exp.  4)  with  HNOs-  A  white  precipitate  of  AgCl  is 
obtained,  owing  to  the  removal  of  NHs  in  the  reversible 
action  above. 


OUTLINE  OF  ANALYSIS  (TABLE  I) 


31 


TABLE  I 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  I 
Ions  present  in  acid  solution 


Reagent 

Bi+++ 

Sb+++ 

Pb++ 

Hg+ 

Ag+ 

I   NH4C1 

BiOCl  (?) 

SbOCl  (?) 

PbCl2 

Hg2Cl2 

AgCl 

2  HCHaN) 

BiCl3 

SbCl3 

PbCl2 

Hg2Cl2 

AgCl 

H2S 

B2iS3  * 

Sb2S3  * 

1 

1 

~T 

3  H20  (Hot) 
(a)   K2Cr2O7 

?l*tl< 

Qrt&ny^ 

PbCl2 
(a)   PbCr04tf  , 

Hg2Cl2 

AgCl 
1 

(&)    H2S04 
4  NH4OH 

(fe)    PbS04 

HgNH2Cl  +  Hg 

Ag(NH3)2Cl 

5  Hn03 

B^  c  /C. 

AgCl 

*  May  be  added  to  Group  II  ppt.  or  tested  according  to  method  outlined  in  Group  II. 


ANALYSIS 

Group  I 


,  Pb++,  Hg+,  Ag+ 

(10)  Precipitation.  —  To  about  25  cc.  of  the  solution 
of  the  substance  acidified  with  HNOs  (see  Discussion  n) 
add  NIHUCl  solution  as  long  as  a  precipitate  continues  to 
form.  Mix  thoroughly  and  allow  the  mixture  to  stand 
for  two  or  three  minutes.  Filter  and  treat  the  precipitate 
by  (n).  Reserve  the  filtrate  for  analysis  of  Group  II. 

(20). 

(n)  Separation  of  Bismuth  and  Antimony.  —  Pour  re- 
peatedly through  the  filter  (10)  a  cold  10  cc.  portion  of  2N. 
HC1.  Treat  the  residue  by  (12).  Dilute  the  filtrate  with 
an  equal  volume  of  water  and  saturate  with  H2S.  A 
brown  precipitate  indicates  the  presence  of  bismuth.  An 
orange-red  precipitate  indicates  the  presence  of  antimony 
and  the  absence  of  bismuth.  Confirmatory  tests  may 
be  made  according  to  the  methods  outlined  in  Group  II, 
or  the  precipitate  may  be  added  to  that  of  the  Group  II 
xsulphide  (20)  and  analyzed  according  to  the  general  scheme. 

(12)  Separation  and  Detection  of  Lead.  —  Pour  re- 
peatedly through  the  filter  (n)  a  10  cc.  portion  of  boiling 


;, 


32  CATIONS— (METAL  IONS) 

water.  Cool,  acidify  with  HC2H3O2  and  add  a  slight 
excess  of  K2Cr2O7  solution.  The  formation  of  a  yellow 
crystalline  precipitate  shows  the  presence  of  lead.  If 
lead  has  been  found,  wash  the  residue  left  by  the  hot 
water,  and  treat  it  by  (13).  (See  Discussion  14.) 

(13)  Detection  of  Mercury. — Pour  repeatedly  through 
the  filter   (12)   a  5-10  cc.   portion  of  NH4OH.     A  black 
residue   left   on    the    filter   paper   shows   the   presence   of 
mercury.     Treat  the  filtrate  by  (14). 

(14)  Detection  of  Ag. — Acidify  the  filtrate   (13)  with 
HNOs.     The  formation  of  a  white  curdy  precipitate  undis- 
solved  by  the  HNOs  shows  the  presence  of  Ag.     (See  Dis- 
cussion 15.) 

DISCUSSION 

11.  The    chlorides  of    lead,   silver    and   mercury  (ous) 
are  insoluble  in  acid  of  moderate  concentration  but  tend  to 
redissolve  in  strongly  acid  solutions,  owing  to  the  forma- 
tion  of   complex   ions   such   as   AgQ2~.     If   the   solution 
to  be  tested  is  strongly  acid  with  an  unknown  acid  it  should 
be  nearly  neutralized  with  NH4OH  before    the  addition 
of   NFUCl.     Complete  neutralization   causes  the   precipi- 
tation of  metal  hydroxides  and  oxy-compounds  which  may 
not  readily  dissolve  on  the  subsequent  addition  of  acid. 
It  is  preferable,   therefore,   not  to  completely  neutralize 
at  this  time. 

12.  Owing  to  the  rather  high  solubility  product  of  PbCl2 
it    is    often    incompletely    precipitated    in    Group    I.     Its 
solubility,  however,  is  considerably  lessened  by  the  addi- 
tion of  a  large  excess  of  NHUCl  becau.se  of  the  common 
ion  effect.     The  mass-law  equation  representing  the  solu- 
bility product  of  PbCl2  is  CPb++  •  C2cr  =K.    Since  the  prod- 
uct of   the  concentrations  of   its  ions  must  under  all  con- 
ditions have  a  definite  value  it  is  readily  seen  that  an 
increase  of   Cl~  must  result  in  a  decrease  of  Pb++.     The 
Pb++,  however,  can  decrease  only  by  forming  non-ionized 
PbCl2,  and  if  the   solution  is  saturated  with   respect   to 


DISCUSSION  (11-15)  33 

PbCl2  molecules,  precipitation  must  take  place.     The  reac- 
tion may  be  represented  by  the  following  equilibrium: 

PbCl2  ^  PbCl2  —  Pb++  +2C1- 

13.  As    was    pointed    out    in    the    preliminary   experi- 
ments, the  precipitation  of  bismuth  and  antimony  in  this 
group  is  due  to  hydrolysis.     Application  of  the  Law   of 
Mass  Action  to  this  reaction  shows  that  their  precipitation 
may  be  practically  or  wholly  prevented  not  only  by  an 
excess  of  acid  but  also  by  an  excess  of  chloride  ion.     For 
the  solubility  product    of  BiOCl  we    have  the  expression 
CBi+++-C0 — •Cci~=K.     The  excess  of  H+ from  the  acid 
unites  with  the  O —  to  form  the  very  slightly  ionized  H2O 
and  so  reduces  the  concentration  of  O — .     This  in  turn 
tends  to  reduce  the  ion  product    and    hence    causes   the 
Bi  to  remain   in   solution.      An  excess  of  Cl~  would  tend 
to   shift   the   equilibrium   toward    the   formation   of    non- 
ionized  Bids  and  so  reduce  the  concentration  of  Bi+  ++. 
The  result  again  is  to  reduce  the  ion  product  and  prevent 
precipitation.     The  same  reasoning  applies  to  the  formation 
of  SbOCl.     (The  student  should  make  the  application.) 

14.  The  solvent  action  of  hot  water  on  PbCl2  is  some- 
what   slow,    hence    the    necessity    of    pouring    the    water 
repeatedly  through  the  filter  containing  the  mixed  chlorides. 
Any  undissolved  PbCl2  left  on  the  filter  may  react  with 
the   NH4OH,    subsequently  added   for  the   separation   of 
mercury  and  silver,   to  form  a  basic  chloride,    PbOHCl. 
This  causes  a  turbidity  in  the  filtrate  containing  the  dis- 
solved  silver.     The  presence  of  this,   however,   does  not 
interfere  with  the  test  for  silver,  since  it  is  readily  dis- 
solved by  the  HNO3. 

15.  The    presence    of    a    large    quantity    of    mercury 
renders  the  test  for  silver  less  delicate,  owing  to  the  reduc- 
ing action  of  free  mercury  on  the  silver  salt. 

Hg+2AgCl^HgCl2+2Ag 
The  silver  thus  reduced  is  insoluble  in  NIHUOH  and 


34  CATIONS— (METAL  IONS) 

is,  therefore,  left  in  the  residue  mixed  with  the 
and  free  mercury.  In  case  the  quantity  of  silver  is  relatively 
small  as  compared  to  the  mercury  it  might  be  completely 
reduced  and  hence  remain  in  the  residue.  If  a  heavy 
black  residue  is  obtained  on  the  addition  of  NH4OH  and 
silver  is  not  found  in  the  nitrate,  the  residue  should  be 
tested  for  silver  as  follows:  Dissolve  the  residue  in  a  small 
quantity  of  aqua  regia,  dilute  with  water  and  filter  off 
the  dissolved  mercury.  Leach  the  residue  with  NH4OH 
to  dissolve  any  AgCl,  and  test  the  filtrate  for  silver  as 
outlined  in  Analysis  (14). 

GROUP  II 
Preliminary  Experiments 

Group  II,  Copper  Division 
Hg++,  Pb++,  Bi+++,  Cu++,  Cd++ 

In  connection  with  the  following  experiments  study 
Tables  II  and  III. 

Experiment  6. — Dilute  separate  5  cc.  portions  of  each 
of  the  test  solutions  containing  the  above  ions  with  an 
equal  volume  of  water  and  saturate  with  H^S.  Allow  the 
precipitates  to  settle;  then  decant  off  the  supernatant 
liquid  (filter  if  necessary)  and  wash  each  of  the  precipitates 
by  decantation  with  about  10  cc.  of  water  in  order  to 
remove  the  remaining  acid.  Write  equations. 

Experiment  7. — To  each  of  the  above  precipitates 
(Exp.  6)  add  10  cc.  of  2N.  HNO3  and  heat  to  boiling. 
Note  the  insolubility  of  HgS.  (Difference,  separation  of 
Hg.)  Write  equations. 

NOTE. — If  too  strong  HN03  is  added  or  if  the  boiling  is  long  continued 
the  HgS  is  largely  converted  into  a  white  difficultly  soluble  substance  of 
the  probable  formula  2HgS-Hg(N03)2  while  some  may  be  completely 
dissolved  as  Hg(N03)2. 


PRELIMINARY  EXPERIMENTS  35 

Experiment  7a. — Pour  off  the  liquid  above  the  residue 
of  HgS  (Exp.  7)  and  add  5  cc.  of  aqua  regia  (3  parts  of 
I2N-HC1  to  i  part  of  I6N-HNO3)  and  warm  gently  until 
solution  is  complete.  Evaporate  almost  to  dryness,  add 
5-10  cc.  of  water  and  then  stannous  chloride  (SnCy 
drop  by  drop.  Write  equations. 

NOTES. — The  failure  of  the  HgS  to  dissolve  in  2N.  HN03  is  due  not 
only  to  the  small  number  of  its  ions  in  a  saturated  solution  but  also  to 
the  slow  removal  of  the  sulphide  ion  by  oxidation  with  the  dilute  HNO3. 
Aqua  regia  is  a  much  stronger  oxidizing  agent.  It  readily  oxidizes  the 
sulphide  ion  and  so  brings  the  mercury  into  solution. 

HgCl2  is  reduced  to  white  Hg2Cl2  by  the  action  of  SnCl2.  This  is  then 
further  reduced  to  free  Hg  by  an  excess  of  SnCl2,  hence  the  change  in  color 
from  white  to  gray. 

Experiment  8. — To  each  of  the  solutions  containing 
Pb++,  Bi+++,  Cu++,  and  Cd++  (Exp.  7)  add  i  cc.  cone. 
H2SO4  and  evaporate  until  the  white  fumes  of  SOs  begin 
to  appear.  Cool  and  add  10  cc.  of  water.  Note  the  fact 
that  PbSCU  remains  insoluble.  (Difference,  separation  of 
Pb.)  Write  equations. 

NOTES. — PbS04  is  somewhat  soluble  in  HN03.  In  order  to  effect  a 
complete  precipitation  of  lead  it  is  necessary  to  remove  the  HNO3.  This 
may  be  done  by  evaporation  since  the  boiling-point  of  HN03  (120.5°) 
is  so  much  lower  than  that  of  the  H2SO4  (338°).  The  appearance  of  the 
white  fumes  of  SO3  is  evidence  of  the  complete  removal  of  HN03. 

Bi2(SO4)3  ordinarily  dissolves  on  the  addition  of  water  but  if  much 
bismuth  is  present  a  coarsely  crystalline  precipitate  of  (BiO)2S04  may 
separate  out  slowly  when  cold,  and  more  quickly  when  heated. 

Experiment  8a. — Decant  or  filter  off  the  clear  liquid 
from  the  insoluble  PbSCX  and  dissolve  it  by  shaking  with 
10  cc.  of  ammonium  acetate  (NH4C2H3O2).  Acidify  the 
solution  with  H^HsCb  and  add  a  few  drops  of  K^Q^Oz. 
The  precipitate  is  PbCrO4.  Write  equations. 

NOTE. — PbS04  is  soluble  in  NH4C2H302  owing  to  the  formation  of  the 
very  slightly  ionized  Pb(C2H3O2)2.  It  should  be  remembered  that  most 
salts  are  highly  ionized.  (See  table,  Introduction,  9.) 


36  CATIONS— (METAL  IONS) 

Experiment  9. — To  each  of  the  solutions  containing 
Bi+  ++,  Cu+  +  ,  and  Cd  +  +  (Exp.  8)  slowly  add  NH4OH  to 
distinctly  alkaline  reaction.  Note  the  formation  of  a 
permanent  white  precipitate  of  BiOOH.  The  precipitates 
of  Cu(OH)2  and  Cd(OH)2  redissolve  on  the  addition  of 
an  excess  of  the  reagent.  (Difference,  separation  of  Bi.) 
Write  equations. 

NOTE.— According  to  the  Law  of  Mass  Action  the  limit  of  solubility 
of  any  acid,  base  or  salt  is  reached  when  the  product  of  the  concentra- 
tion of  its  ions  equals  a  certain  maximum,  called  the  solubility  product. 
(Cone,  of  pos.  ions X  Cone,  of  neg.  ions  =  a  const.,  solubility  product.) 
If  NH4OH  is  added  to  a  bismuth  salt,  precipitation  of  BiOOH  begins 
when  enough  OH~  have  been  added  to  reach  the  solubility  product  for 
BiOOH.  The  greater  the  concentration  of  OH~,  i.e.,  the  greater  the 
excess  of  NH4OH,  the  smaller  must  be  the  concentration  of  the  Bi+  +  + 
and  hence  the  more  complete  the  precipitation.  It  will  be  recalled, 
however,  that  an  addition  of  an  excess  of  NH4OH  dissolves  both  Cu(OH)2 
and  Cd(OH)2.  This  is  explained  by  the  fact  that  the  Cu++  and  Cd+  + 
unite  with  HN3  to  form  the  complex  ions  Cu(NH3)4++  and  Cd(NH3)4+  + 
(See  Introduction  15.)  This  removes  the  Cu++  and  Cd++  as  such 
and  prevents  their  precipitation.  The  Bi  +  ++,  incapable  of  forming  a 
complex  ion  with  NH3,  must  be  removed  by  precipitation  as  indicated 
above. 

Experiment  pa. — Decant  or  filter  off  the  clear  liquid 
from  the  precipitate  of  BiOOH  and  dissolve  it  in  a  few 
cc.  of  6N.  HC1.  Evaporate  to  about  I  cc.  and  pour  into 
a  large  volume  of  warm  water.  Filter  off  the  white  milky 
precipitate  of  BiOCl  and  pour  over  the  precipitate  on  the 
filter  paper  a  freshly  prepared  solution  of  sodium  stannite 
(Na2$nO2).  (See  Appendix  i.)  The  black  residue  is 
finely  divided  bismuth.  Write  equations. 

NOTE. — The  solution  of  BiCl3  is  evaporated  to  remove  the  excess 
of  HC1  in  order  to  promote  hydrolysis.  (See  Preliminary  Exp.  i,  also 
Introduction,  14.) 

Experiment  10. — Divide  each  of  the  solutions  contain- 
ing copper  and  cadmium  (Exp.  9)  into  two  parts.  Acidify 


PRELIMINARY  EXPERIMENTS  37 


the  first  portion  of  each  with  H^HsCb  and  add  a  few 
drops  of  K4Fe(CN)6.  Note  the  color  of  the  precipitates 
formed.  (Difference,  detection  of  Cu.)  Write  equations. 
To  the  second  portion  containing  copper  add  KCN 
(Care,  Poison)  until  the  blue  color  just  disappears;  then 
add  an  equal  volume  to  the  second  portion  containing 
cadmium.  Saturate  each  with  H^S.  Note  the  formation 
of  insoluble  CdS.  (Difference,  detection  of  Cd.) 

NOTE.  —  Excess  KCN  reacts  with  salts  of  copper  and  cadmium  to 
form  the  complex  ions  Cu(CN)2++  and  Cd(CN)4++.  The  Cu(CN)4+  + 
first  formed  immediately  decomposes  into  Cu(CN)2+  and  C2N2.  The 
C2N2  reacts  with  NH4OH  to  form  NH4CNO  and  other  more  or  less  com- 
plex substances.  The  failure  to  obtain  a  precipitate  with  H2S  is  evidence 
that  not  enough  Cu+  or  Cu++  is  present  to  reach  the  solubility  product 
for  Cu2S  or  CuS  and  shows  that  the  complex  ion  Cu(CN)2+  is  only  very 
slightly  dissociated.  On  the  other  hand,  the  complex  ion  Cd(CN)4+  + 
must  be  dissociated  to  a  considerable  extent  since  it  produces  sufficient 
Cd++  to  reach  the  solubility  product  for  CdS.  The  reactions  proceed 
as  follows: 

2Cu(NH3)4SO4  +4KCN  -*  2KCu(CN)2  +C2N2  +4NH3  +K2SO4 
Cd(NH3)4S04+4KCN^  K2Cd(CN)4+4NH3  +K2SO4 


CATIONS— (METAL  IONS) 


TABLE  II  t 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  OF  GROUP  II 

(SEPARATION  INTO  Cu  AND  SN  DIVISIONS) 

Ions  present  in  O.3N.  HC1  solution 


As  + 

Sn++ 

No. 

Reagent 

Hg++ 

Pb++ 

Bi   + 

Cu++ 

Cd++ 

or 

+  + 
Sb  + 

or 

As  +  + 

Sn  +  + 

i 

H2S 

HgS 

PbS 

Bi2S3 

CuS 

CdS 

As2S3  As2S5 

Sb2S3 

SnS  or  SnS2 

2 

(NH4)2Sx 

HgS 

PbS 

Bi2S3 

CuS 

CdS 

(NH4)3AsS4 

(NH4)3SbS4 

(NH4)2SnS3 

3 

6N.  HC1 

1 

1 

I 

I 

1 

As2S5 

Sb2S5 

SnS2 

I 

I 

1 

ANALYSIS 
Group  II 

Hg++,  Pb++,  Bi  +  ,  Cu++,  Cd++, 


As  +  ,  As+  +  ,  Sb  +  ,  Sn++,  Sn 


+- 


(20)  Precipitation.  —  Dilute    the     filtrate    (10)    which 
should  contain  just  5  cc.  of  6N.  acid  (see  Discussion  16) 
to  100  cc.  and  saturate  with  H2S.     Filter,  heat  the  filtrate 
to  boiling  and  again  saturate  with  I-bS. 

If  no  further  precipitate  forms  reserve  the  solution  for 
analysis  of  Group  III  (50)  and  treat  the  precipitate  by  (21). 

If  a  further  precipitate  forms  (see  Discussion  17) 
evaporate  the  mixture  to  dryness,  moisten  the  residue 
with  cone.  HC1  and  evaporate  to  dryness  again  to  remove 
all  HNO3.  Add  10-15  cc.  of  6N.  HC1,  heat  to  boiling 
and  pass  H2S  into  it  for  five  to  ten  minutes.  Filter,  unite 
the  precipitate  to  that  above  and  treat  by  (30).  Reserve 
the  filtrate  for  analysis  of  Group  III  (50). 

(21)  Separation  of  the  Copper  and  Tin  Divisions.— 
Transfer  the  precipitate  (20)  to  an  evaporating  dish,  add 
10—15  cc.  of  (NHU^Sz  reagent,  cover  the  dish  with  a  watch 
glass  and  warm  the  mixture  for  about  five  minutes.     (Do 
not   boil.)     (See   Discussion    18.)     Dilute   with   an   equal 


DISCUSSION  (16-20)  39 

volume  of  water  and  filter.  A  second  treatment  with 
(NH^Sz  reagent  should  be  made  if  the  residue  is  large 
and  much  was  extracted  by  the  first  treatment.  Wash  the 
residue  with  hot  water  (see  Introduction  3,  /)  and  treat  it 
by  (30).  Treat  the  filtrates  separately  by  (22). 

(22)  Precipitation  of  the  Tin  Division.  —  To  the  first 
filtrate  obtained  on  treating  the  Group  II  sulphides  with 
(NH4)2Sz  reagent  (21)  add  6N.  HC1  with  frequent  stirring 
until  the  solution  remains  milk-white  from  the  separation 
of  finely  divided  sulphur.  Allow  the  mixture  to  stand  for  one 
or  two  minutes  to  coagulate  the  precipitate.  The  presence  of 
the  tin  division  is  indicated  by  the  presence  of  a  flocculent 
yellow  or  orange  precipitate.  (See  Discussion  19.)  Treat  the 
second  filtrate  obtained  in  (21)  in  the  same  way,  and  if  a 
flocculent  precipitate  forms,  unite  it  with  the  one  obtained 
above.  Filter  and  wash  the  precipitate,  using  suction  to 
dry  it  as  thoroughly  as  possible.  (See  Discussion  20.) 
Reject  the  filtrate  and  treat  the  precipitate  by  (40). 

DISCUSSION 

1  6.  The  precipitation  of  the  sulphides  of  the  metals 
of  both  Groups  II  and  III  is  determined  very  largely  by 
the  acid  concentration  of  the  solution.  From  the  stand- 
point of  the  Law  of  Mass  Action  and  the  Ionic  Theory 
two  factors  must  be  taken  into  consideration  in  the  pre- 
cipitation of  the  metal  sulphides,  viz.,  the  value  of  the 
solubility  product  and  the  concentration  of  the  sulphide 
ion.  The  solubility  product  varies  with  the  nature  of  the 
sulphide  and  with  the  temperature.  The  order  of  pre- 
cipitation with  H2S  from  cold  HC1  solution  of  decreasing 


acid    concentration    is    about    as    follows:    As  ++  ,  As  +  , 

Hg++,  Cu++,  Sb+++,  Bi+++  and  Sn++,    Cd++,    Pb++   and 
Sn++,  Zn++,  Fe++,  Ni++,  Co++,  Mn++. 

According  to  Henry's  Law  of  the  Solubility  of  Gases, 
the  solubility  and  therefore  the  concentration  of  H2S 
will  have  a  constant  value  at  a  definite  temperature  and 


40  CATIONS—  (METAL  IONS) 

pressure.     H^S   ionizes  to  a  slight    extent  into   H+   and 
HS~  and  to  a  much  less  extent  into  2H+  and  S=.     Since 
the  S=  is  the  active  agent  in  the  precipitation  under  dis- 
cussion only  the  latter  ionization  need  be  considered  here. 
The  expression  of  the  Law  of  Mass  Action  for  IHUS  is 

C2H+-CS= 
—    -     -  =K 


In  a  saturated  solution,  however,  CH2s  is  a  constant; 
hence  simplifying  the  above  equation  gives 

C2H+-Cs==K.CH2s=K 

From  this  it  is  readily  seen  that  any  increase  in  the  con- 
centration of  H+  must  result  in  a  corresponding  decrease 
in  S=.  But  in  order  that  a  sulphide,  e.g.,  CdS,  may 
precipitate,  its  solubility  product,  CCd++-Cs==  =Kcds,  must 
be  reached.  Therefore,  if  the  acid  concentration  is  large 
the  concentration  of  S=  may  become  so  small  that  its 
product  with  CCd++  will  not  equal  the  solubility  product 
for  CdS.  On  the  other  hand,  if  the  acid  concentration 
is  small  the  concentration  of  S=  may  become  so  large  that 
its  product  with  CZn++  will  equal  the  solubility  product 
for  ZnS,  in  which  case  ZnS  would  be  precipitated  in 
Group  II.  In  order  to  make  a  complete  separation 
between  Groups  II  and  III,  the  concentration  of  S=  must 
be  so  regulated  that  there  will  be  sufficient  to  reach  the 
solubility  product  of  the  most  soluble  of  the  sulphides  of 
Group  II  but  not  sufficient  to  reach  the  solubility  product 
of  the  least  soluble  of  Group  III.  (See  table,  Introduc- 
tion n).  The  necessity  of  following  the  directions  very 
accurately  is  therefore  evident. 

17.  When  H2S  is  passed  into  a  cold  acid  solution  con- 
taining an  arsenate,  reaction  takes  place  only  very  slowly 
between  the  two  substances.  Part  of  the  arsenic  is  pre- 
cipitated as  As2S5  and  a  part  is  reduced  to  the  trivalent 
state  and  precipitated  as  As2Ss.  When  the  solution  is 
heated  the  reduction  takes  place  somewhat  more  rapidly. 
The  most  favorable  conditions  for  the  reduction  and  pre- 


DISCUSSION  (16-20)  41 

cipitation  of  arsenates  with  H2S  is  from  a  hot  solution 
having  a  comparatively  high  concentration  of  acid. 

1 8.  In   the   separation   of   arsenic,    antimony   and   tin 
from  the  copper  division,     advantage  is  taken  of  the  acid 
character  of  the  elements  of   the  tin  division  and  their 
ability  to  form  sulpho-salts  with  (NH4)2Sz.     This  reagent, 
however,  has  a  slight  solvent  action  on  the  sulphides  of 
copper  and  mercury.  If  (NH4)2Sz,  to  which  about  5  per  cent 
NaOH  has  been  added,  is  used,  the  solvent  action  on  the 
sulphides  of  copper  and  mercury  is  very  much  reduced  and  a 
more  satisfactory  separation  is  obtained.     Excessive  heat 
not  only  increases  the  solvent  action  on  HgS  and  CuS  but 
decomposes  the  reagent  with  precipitation  of  free  sulphur. 

19.  If   the   separation   of   arsenic,    antimony   and    tin 
from  the  copper  division  has  been  complete  the  presence 
of  the  tin  division  is  shown  by  a  flocculent  yellow  or  orange 
precipitate  on  acidification  of  the  (NH^S*  solution  with 
HC1.     However,    if    the    precipitate    obtained    is    brown 
(indicating  copper)  dark  gray  (indicating  mercury),  or  of 
unpronounced  yellow  or  orange,  so  as  to  make  doubtful  the 
presence  of  the  tin  division,  much  time  can  often  be  saved 
by  proceeding  as  directed  below.    Small  amounts  of  copper 
which  otherwise  might  be  overlooked  may  also  be  detected. 

Heat  the  precipitate  with  10-20  cc.  of  NH4OH  almost 
to  boiling  for  five  minutes,  and  filter.  This  treatment 
dissolves  all  but  the  copper  and  free  sulphur.  The  residue, 
therefore,  should  be  tested  for  copper  according  to  (31) 
(34)  and  (36)  unless  copper  has  already  been  found.  Pass 
H2S  into  the  filtrate  for  fifteen  to  twenty  seconds  to  pre- 
cipitate any  HgS  and  change  the  partially  sulphurated 
acids  (NH4)3AsO3S,  (NH4)3AsO2S2,  etc.,  into  the  fully 
sulphurated  form  (NH4)sAsS4,  etc.  Filter  if  necessary 
and  acidify  the  filtrate  with  HC1.  Filter  off  the  precipi- 
tated sulphides,  wash,  dry  by  suction  and  treat  by  (40). 

20.  If  a  suction  pump  is  not  available  the  precipitate 
may  be  dried  satisfactorily  by  pressing  the  filter  containing 
it  between  several  thicknesses  of  clean  filter  paper. 


42 


CATIONS— (METAL  IONS) 


I 

H 

Q 

Q 
fc 

5  §1 

3   <% 
«   §5 


u 


uu 


I 

S 


3    SI 


<-s          U 

CU         PL, 


! 


ii 


c|§ 


OUTLINE  OF  ANALYSIS  (TABLE  III)  4? 

ANALYSIS 
Group  II,  Cu  Division 

(30)  Separation    of    Mercury. — Transfer    the    residue 
(21)  to  a  porcelain  dish  and  add  10-20  cc.  of  2N.  HNOs 
and  heat  to  boiling.     Boil  gently  for  two  to  three  minutes, 
not  longer.     (See  note,  Preliminary  Exp.  7).     Filter,  wash 
the  residue  and  treat  it  by  (31).     Treat  the  filtrate  by  (32). 

(31)  Confirmatory    Test   for    Mercury. — Transfer    the 
residue    (30)    undissolved   by   HNOs   to  a  porcelain  dish 
and  add  5-10  cc.  of  aqua  regia  (see  Preliminary  Exp.  7a). 
Warm   gently   till   solution   is   complete,    then   evaporate 
almost  to  dryness,   dilute  with  5-10  cc.   of  water,   filter 
and  add  to  the  clear  filtrate  some  SnCl2  solution,  at  first 
1-2  drops  then  2-3  cc.  (see  Discussion  22).     The  formation 
of  a  white  precipitate  which  turns  gray  on  the  addition 
of   excess   SnCb   shows   the   presence   of   mercury.     (See 
Discussion  23.) 

(32)  Separation  of  Lead. — To  the  filtrate  obtained  in 
(31)  add  2-3  cc.  of  cone.  H2SO4,  transfer  to  a  porcelain 
dish  and  evaporate  until  the  dense  white  fumes  of  SOs 
appear.     Cool,    pour    the    mixture    into    a    small    beaker 
containing  10—15  cc.  of  water  and  rinse  out  the  vessel  with 
a  portion  of  the  solution  formed,  in  order  to  be  sure  that 
all  of  the  solid  is  transferred.     Allow  the  mixture  to  stand 
four  to   five   minutes.     The  formation  of  a  fine,   white, 
crystalline    precipitate    indicates    the    presence    of    lead. 
The  precipitate  may  be  more  readily  distinguished  if  the 
liquid  is  given  a  slight  whirling  motion,  so  that  the  pre- 
cipitate will  collect  on  the  bottom  of  the  beaker  toward 
the  center.     Filter,  treat  the  precipitate  by  (33)  and  the 
filtrate  by  (34). 

(33)  Confirmatory  Test  for  Lead. — Dissolve  the  pre- 
cipitate of  PbSC>4  (32)  by  pouring  a  10-20  cc.  portion  of 
ammonium    acetate    (NH4C2H3O2)    solution    repeatedly 
through  the  filter.     To  the  filtrate  add  a  few  drops  of 


44  CATIONS— (METAL  IONS) 

K2Cr2O7  and  3-5  cc.  of  HC2H3O2.     A  yellow  precipitate 
is  PbCrO4  (see  Discussion  24). 

(34)  Separation  of  Bismuth. — To  the  filtrate  obtained 
in  (32)  add  NH4OH  until,  after  shaking,  a  distinctly  alka- 
line   reaction    is    obtained.     (Test    with    litmus.)     Shake 
to  coagulate  the  precipitate  of  BiOOH,  and  filter.  Wash  the 
precipitate  and  treat  it  by  (35).     Treat  the  filtrate  by  (36). 

(35)  Confirmatory  Test  for  Bismuth. — Pour  through 
the  filter  containing  the  precipitate  of  BiOOH  (34)  a  cold 
freshly  prepared  solution  of  Na2SnO2  (see  Appendix  I). 
The  formation  of  a  black  residue  shows  the  presence  of 
bismuth.     (See  Discussion  25.) 

(36)  Detection  of   Copper. — If  the   filtrate  from   the 
BiOOH  (34)  is  deep  blue,  copper  is  shown  to  be  present. 
If,  however,  it  is  colorless  or  nearly  so,  about  one-fourth 
of  the  solution  should  be  acidified  with  HC2H3O2  and  a 
few  drops  of  K4Fe(CN)6  added.     The  formation  of  a  red 
precipitate  or  coloration  shows  the  presence  of  copper. 
Treat  the  remainder  of  the  solution  by  (37)  or  (38.) 

(37)  Detection   of   Cadmium. — To   the   remainder   of 
the  NH4OH  solution  obtained  in  (36)  add  KCN  solution 
(care,  poison)  until  the  blue  color  just  disappears;    if  the 
solution  is  colorless  add  only  a  few  drops.     Pass  H2S 
into  the  colorless  solution  for  about  half  a  minute.     The 
formation  of  an  immediate  yello>w  precipitate  (see  Dis- 
cussion 26)  shows  the  presence  of  cadmium. 

(38)  Optional  Method  for  the  Detection  of  Cadmium. — 
Acidify  the  remainder  of  the  NH4OH  solution  (36)  with 
H2SO4,  add  a  few  iron  nails  or  some  iron  filings  and  boil 
for  a  short  time.     Filter,  and  unless  the  solution  is  still 
acid  make  it  just  acid  with  6N.H2SO4  and  pass  H2S  into 
it.     The  formation  of  a  yellow  precipitate  shows  the  pres- 
ence of  cadmium. 

Discussion 

21.  If  the  elements  of  the  copper  division  are  present 
in  large  quantity,   small   quantities  of  tin   may  remain 


DISCUSSION  (21-27)  45 

undissolved  by  the  (NH4)2Sz  treatment.  This  may  also 
occur  when  small  quantities  of  cadmium  and  stannous 
tin  are  present  together.  Any  tin  sulphide,  either  SnS 
or  SnS2,  remaining  in  the  copper  division,  will  be  con- 
verted by  the  HNOs  (30)  into  insoluble  metastannic  acid, 
H2SnOs.  This  is  practically  unaffected  by  the  aqua 
regia  or  bromine  water  used  in  (31). 

In  order  to  recover  any  tin  that  may  be  left  in  the  copper 
division  the  residue  obtained  by  treatment  with  aqua 
regia  (31)  should  be  treated  as  follows:  If  it  is  dark- 
colored,  showing  incomplete  removal  of  HgS,  digest  with 
bromine  water  in  order  completely  to  remove  the  HgS, 
filter  and  dissolve  the  residue  in  a  small  quantityof(NH4)2Sa; 
reagent.  Dilute,  filter  if  necessary,  and  add  the  solu- 
tion to  the  main  ammonium  sulphide  solution  obtained 
in  (21). 

22.  The  addition   of  SnCl2   to  a  solution   of   HgCl2 
causes    an    immediate    reduction    to    the    white    Hg2Cl2. 
Further  addition  of  SnCl2  carries  the  reduction  to  free 
mercury,  which  imparts  a  gray  appearance  to  the  pre- 
cipitate.   The  latter  reduction  is  hindered  very  materially 
and  may  be  almost  prevented  if  excess  aqua  regia  has 
not   been   completely   removed   before   making   the   test. 
Bromine  water  may  be  used  in  place  of  the  aqua  regia. 
Excess  bromine,   however,   must  be  removed  before  the 
addition  of  SnCk. 

23.  If  the  separation  of  mercury  is  incomplete  and  a 
large  amount  of  CuS  is  left  undissolved  by  the  HNOs 
treatment  (30),  a  white  precipitate  of  CuCl  will  separate 
out  on  the  addition  of  SnCl2.     This,  however,  does  not 
turn  gray  with  excess  SnCl2. 

24.  The  confirmatory  test  for  lead  should  always  be 
made,   since   a   precipitate   with   H2SO4   may   consist   of 
(BiO)2SO4  or  BaSO4.     The  (BiO)2SO4  is  coarsely  crystal- 
line, dissolves  in  NH4C2H3O2  and  gives  a  yellow  color 
with   K3Cr2O7.     The   precipitate,    however,    differs    from 
PbCrO*  in  that  it  dissolves  readily  in  HC2H3O2.     The 


46  CATIONS— (METAL  IONS) 

BaSO4  resembles   PbSO4  in  appearance  but  is  insoluble 
in  NH4C2H3O2. 

25.  If,   owing  to  occlusion  or  incomplete  washing  of 
the    H2S    precipitate,    an    incomplete    separation     from 
Group   III  is  obtained,   the  addition  of  NH4OH  for  the 
separation    of    bismuth    will    cause    the    precipitation    of 
Fe(OH)3  or  other  hydroxides  of  Group  III.     These  differ 
from  BiOOH  in  that  none  of  them  are  reduced  by  Na2SnC>2 
by  short  contact  in  the  cold. 

26.  An  immediate  yellow  precipitate  with  H^S  shows 
the  presence  of  cadmium.     When  much  copper  is  present 
and  the  solution  is  saturated  with  H^S  a  deep  yellow  color 
soon  develops  and  an  orange-red  precipitate  of  (CSNH2)2 
may  separate  out  on  standing,  owing  to  a  reaction  between 
the  H2S  and  C2N2  set  free  by  the  reduction  of  Cu(CN)2. 

27.  If,  owing  to  previous  errors  in  analysis,  a  black 
precipitate    (due  to   HgS,    PbS,   etc.)    is  obtained   in   the 
final  test  for  cadmium  with  I-bS,  it  should  be  thoroughly 
washed   and   the  cadmium  dissolved  out  by   boiling  the 
precipitate  with  15  cc.  of  1.2 N.  H2SO4  and  filtering.    After 
it  has  been  diluted  with  two  to  three  times  its  volume  of 
water  the  CdS  if  present  may  be  precipitated  by  saturating 
with  H2S. 


Preliminary  Experiments 

Group  II,  Tin  Division 

+  +         +  +  +         ++  +  + 

As  +  ,  As  ++  ,  Sb  +  ,  Sn++,  Sn+- 

In  connection  with  the  following  experiments  study 
Tables  II  and  IV. 

Experiment  n. — Dilute  separate  5-cc.  portions  of  the 
test  solutions  of  each  of  the  above  ions  with  an  equal  volume 
of  water,  and  saturate  cold  with  H2S.  If  a  precipitate 
does  not  form  heat  to  boiling  and  saturate  again  with 


PRELIMINARY  EXPERIMENTS  47 


H2S.     Note  the  difference    in    action  between   As  +    and 


As  ~l~+  .     Write  equations. 

Experiment  12.  —  Decant  or  filter  the  liquid  from  the 
precipitates  (Exp.  n)  and  dissolve  each,  by  warming  if 
necessary,  with  about  5  cc.  of  (NH^S*  reagent.  Do 
NOT  boil 

NOTES.  —  The  solubility  of  the  sulphides  of  arsenic,  antimony  and  tin 
is  doubtless  due  to  their  acidic  character  and  tendency  to  form  sulpho- 
salts.  This  is  in  direct  contrast  to  the  metals  of  the  copper  division 
of  Group  II  in  that  the  latter  are  all  basic  in  character  and  have  very 
little  tendency  to  form  sulpho-salts. 

The  use  of  (NH4)2Sa;  instead  of  (NH2)4S  is  made  necessary  only  in 
the  cases  of  Sb2S3  and  SnS.  Antimony  and  tin  must  be  oxidized  to  the 
higher  valence  since  the  sulpho-salt  of  the  lower  valence  is  apparently 
incapable  of  existence.  The  following  reactions  for  the  solution  of 
As2S3  and  As2Ss  are  as  follows: 


S^  2(NH4)3AsS4  +  (6;x;  -s)S 
As2S5+6(NH4)2Sx->  2(NH4)3AsS4  +  (6JC  -3)S 


Write  the  corresponding  equations  for  the  solution  of  Sb2S3,  SnS  and 
SnS2  noting  the  fact  that  in  the  sulpho-salt  arsenic  and  antimony  are 
quinquivalent  while  tin  is  quadrivalent. 

Experiment  13.  —  Since  the  two  solutions  of  arsenic 
are  now  identical,  (NH^AsS^  as  are  also  those  of  tin, 
(NH4)2SnS3,  one  of  each  may  be  discarded.  Dilute  the 
remaining  solutions  of  arsenic,  antimony  and  tin  (Exp.  12) 
with  an  equal  volume  of  water  and  acidify  with  6N.  HC1. 
Note  the  precipitate  formed  in  each  case.  Write  equations. 

NOTE.  —  When  (NH4)2SX  is  treated  with  an  acid,  considerable  sulphur 
is  apt  to  be  set  free.  This  gives  a  white,  milky  appearance  to  the  mixture 
and  may  prevent  the  detection  of  small  amounts  of  the  sulphides.  Dilu- 
tion with  water  obviates  this  difficulty  to  some  extent. 


Experiment  14.  —  To  each  of  the  precipitates  of 
Sb2S5  and  SnS2  (Exp.  13)  dried  by  suction  or  by  pressing 


48  CATIONS— (METAL  IONS) 

between  filter  paper,  add  just  10  cc.  of  I2N.  HC1.  Place 
the  test-tubes  containing  the  'mixtures  in  a  beaker  of 
water  and  heat  to  boiling.  Boil  gently  for  ten  minutes. 
Note  the  solution  of  Sb2S5  and  SnS2.  (Difference,  sepa- 
ration of  As.)  Write  equations. 

NOTES. — If  the  boiling  is  carried  out  so  that  the  bubbles  rise  but  slowly 
from  the  solutions,  practically  no  As2S6  is  dissolved.  However,  if  the 
boiling  is  too  vigorous  some  of  the  arsenic  may  go  into  solution.  This 
is  due  to  removal  of  H2S  and  consequent  shifting  of  equilibrium. 

Sb2S5  dissolves  in  i2N.  HC1  somewhat  slowly,  especially  when  large 
amounts  are  present.  It  dissolves  with  the  formation  of  SbCl3,  the  anti- 
mony being  reduced  by  the  H2S  to  the  trivalent  condition, 

Experiment  i4a. — Without  filtering,  add  small  particles 
of  KClOs  or  NaClOa,  a  few  at  a  time,  to  the  As2S5  and 
stir  till  solution  is  complete.  Evaporate  to  1-2  cc.  Make 
distinctly  alkaline  with  NH4OH  and  add  1-2  cc.  of 
magnesia  mixture  (MgCl2-NH4Cl-NH4OH).  If  no  pre- 
cipitate forms  add  about  one-fourth  its  volume  of  NIrUOH 
and  rub  the  inside  of  the  tube  with  a  glass  rod  to  hasten 
precipitation. 

NOTE.— The  action  of  HC1  on  KC103  or  NaC103  is  to  liberate  chlorine 
and  C1O2,  which  remove  the  S  =by  oxidation,  causing  the  arsenic  to  go 
into  solution.  It  is  dissolved  as  HsAs04  instead  of  AsCl5,  very  little 
of  which  remains  as  such  even  in  very  concentrated  HC1. 

MgNHiAsCX  is  somewhat  soluble  even  in  cold  water;  therefore  it  is 
necessary  to  have  a  rather  concentrated  solution.  It  also  has  a  tendency 
to  hydrolize  into  NH4OH  and  MgHAs04,  a  soluble  salt.  This  tendency 
is  counteracted  by  the  addition  of  a  large  excess  of  NH4OH.  It  has  an- 
other peculiarity,  which  is  not  uncommon,  in  that  it  is  inclined  to  super- 
saturate. This  may  be  overcome  and  precipitation  effected,  by  vigorous 
shaking,  or  by  producing  a  rough  surface  in  contact  with  the  liquid. 

Experiment  15. — Dilute  the  solutions  of  antimony  and 
tin  (Exp.  14)  to  50  cc.  and  transfer  one- third  of  each  to  a 
third  vessel.  Heat  the  remaining  portions  to  90°  and 
saturate  with  FbS.  Note  the  precipitation  of  Sb2Ss. 
(Difference,  separation  of  Sb.)  Write  equations. 

Experiment    isa. — Dilute    the   solution   in   the   third 


PRELIMINARY  EXPERIMENTS  49 

vessel  (Exp.  15)  to  50  cc.  and  saturate  with  H2S.  Note 
the  color  of  the  precipitate.  Without  filtering,  boil  to 
expel  H2S.  Add  bromine  water  and  heat  till  solution 
is  complete,  then  add  5  gms.  of  solid  oxalic  acid  (H2C2O4) 
and  saturate  with  H2S.  Note  the  precipitation  of  Sb2S3. 
(Difference,  separation  of  Sb.)  Write  equations. 

NOTES. — There  is  enough  difference  in  the  solubility  products  of 
Sb2S3  and  SnS2  for  the  careful  worker  to  make  a  complete  separation  of 
antimony  and  tin  according  to  Exp.  15.  However,  if  the  directions  for 
temperature  and  acid  concentration  are  not  carefully  observed,  a  pre- 
cipitate containing  both  antimony  and  tin  may  be  obtained  (Exp.  150), 
in  which  case  the  separation  can  be  made  according  to  Exp.  1 50. 

The  bromine  water  is  used  to  oxidize  the  S=  and  so  hasten  the  solu- 
tion. The  same  results  may  be  obtained  more  slowly  by  evaporation 
to  concentrate  the  acid. 

The  failure  of  SnS2  to  precipitate  in  the  presence  of  a  large  excess  of 

H2C2O4  is  doubtless  due  to  the  formation  of  a  complex  which  reduces  the 

+  + 
concentration  of  Sn+  +  . 

Experiment  16. — Dilute  the  solution  containing  tin 
(Exp.  15)  to  50  cc.  and  saturate  cold  with  H2S.  Note  the 
precipitation  of  SnS2.  Without  filtering,  add  1-2  gms. 
of  granulated  test  lead  and  boil  two  to  three  minutes. 
Cool  and  filter,  allowing  the  filtrate  to  run  into  a  solution 
of  HgCl2.  The  precipitate  is  Hg2Cl2+Hg. 

+  + 

NOTE. — The  action  of  lead  is  to  reduce  the  Sn++  to  Sn++,  which  in 
turn  will  reduce  HgCl2  to  HgiCl2  or  even  to  free  mercury.  Stannic  tin 
has  no  effect  on  HgCl2;  hence  the  necessity  for  reduction  before  the 
final  test. 


50 


CATIONS— (METAL  IONS) 


TABLE  iy 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  II 

(SN  DIVISION) 


No. 

Reagent 

As2S5 

SbjS, 

SnS2 

i 

I2N.  HC1 

As2S5 

SbCl3 

SnCl4 

HC1 

KC1O3 

H3AsO4 

Magnesia  mixture 

MgNH4AsO4 

6N.  HC1 

H3AsO4 

H2S 

As2S3-5 

2 

H2S 

1 

Sb2S3 

SnCl4 

3 

HCl-Br2 

SbCl3 

SnCl4 

4 

H2C204 

i 

H2S 

Sb2S3 

5 
6 

Pb 
HgCl2 

i- 

SnCl2 
SnCl4(Hg2Cl2) 

ANALYSIS 
Group  II,  Tin  Division 

(40)  Separation  of  Arsenic. — Transfer  the  precipitated 
sulphides  (22),  dried  by  suction  or  by  pressing  between 
filter  paper,  to  a  test-tube,  and  add  just  10  cc.  of  I2N.  HC1. 
(See  Discussion  28.)     Place  the  test-tube  in  a  beaker  of 
water  and  heat  until  the  contents  of  the  tube  just  begin 
to  boil.     Keep  the  water  at  this  temperature  for  about 
ten  minutes,  occasionally  stirring  the  contents  of  the  tube 
with   a  glass   rod.     When   the   reaction   is   complete   and 
no  further  solution  takes  place,  dilute  with  5  cc.  of  water 
and  filter,  allowing  the  filtrate  to  run  into  a  5o-cc.  graduate. 
Wash  the    residue  with  5-10  cc.   of  water,   catching  the 
wash  water  in  the  graduate  with  the  filtrate.     Test  the 
residue  for  arsenic  by  (41)  and  treat  the  filtrate  by  (43). 

(41)  Detection  of  Arsenic. — Punch  a  hole  through  the 
filter  and  wash  the  residue  of  As2S5  into  a  test-tube  with 
5-10  cc.  of  6N.  HC1.     Warm  the  mixture  gently,  adding 
solid  KC1O3  or  NaClO3,  one  crystal  at  a  time,  until  the 


OUTLINE  OF  ANALYSIS  (TABLE  IV)  51 

arsenic  is  dissolved.  Make  the  solution  just  alkaline 
with  NH4OH,  then  add  about  one- third  its  volume  of 
I5N.  NH4OH  and  about  0.5  cc.  of  magnesia  mixture 
(MgCl2-NH4Cl-NH4OH).  Shake  the  contents  vigor- 
ously and  if  no  precipitate  forms  rub  the  walls  of  the  tube 
with  a  glass  rod  and  allow  it  to  stand  for  some  time.  (See 
note,  Preliminary  Exp.  I4a).  The  formation  of  a  white 
crystalline  precipitate  indicates  the  presence  of  arsenic. 
Filter  and  confirm  the  arsenic  by  (42). 

(42)  Confirmatory    Test    for    Arsenic. — Dissolve    the 
precipitate   of    MgNH4AsO4     (41)     by   pouring   5-10    cc. 
of  6N.  HC1  repeatedly  through  the  filter.     Heat  the  solu- 
tion to  boiling  and  pass  H2S  into  it  for  five  to  ten  minutes. 
The  formation  of  a  white  precipitate,  changing  to  yellow, 
shows  the  presence  of  arsenic.     (See  Discussion  17.) 

(43)  Detection  of  Antimony. — To  the  filtrate  (40)  add 
water  just  sufficient  to  make  a  total  volume  of  50  cc. ; 
transfer  to  a  small  flask  and  heat  to  about  90°.     Pass 
H2S  into  the  hot  solution  for  about  five  minutes,  keeping 
the  temperature  at  about  90°  by  placing  the  flask  in  a 
beaker    containing    boiling  water.     The    formation  of  an 
orange-red   precipitate  shows  the  presence   of  antimony. 
(See  Discussion  29.)     Filter  while  hot,  add  5  cc.  of  water, 
heat  to  90°  and  again  saturate  with  H2S  to  remove  all 
antimony.     Filter  if  a  precipitate   forms,   and   treat   the 
filtrate  by  (45). 

(44)  Confirmatory  Test  for  Antimony. — If  the  precipitate 
(43)  is  not  orange,  and  the  presence  of  antimony  is  doubt- 
ful, boil  the  mixture  without  filtering  until  the  precipitate 
has    redissolved.      Add    5-10   gms.    of    solid    oxalic   acid 
(H2C2O4)  and  pass  H2S  into  the  hot  solution.     If  antimony 
is  present  a  bright  red  precipitate  will  be  formed.     Filter 
and  test  the  filtrate  for  tin  by  (46). 

(45)  Detection  of  Tin. — Dilute  the  filtrate  from  (43) 
to  70  cc.,   cool  and  pass  in  H2S  for  ten  minutes.     The 
formation  of  a  yellow  precipitate  indicates  the  presence 
of  tin. 


52  CATIONS— (METAL  IONS) 

(46)  Confirmatory  Test  for  Tin. — Boil  the  mixture 
(45)  or  the  filtrate  from  (44)  with  1-2  gms.  of  granulated 
test  lead  for  two  to  three  minutes.  Cool  the  mixture 
and  filter  into  a  solution  of  mercuric  chloride  (HgCl2). 
The  presence  of  tin  is  shown  by  the  formation  of  a  white 
precipitate,  which  may  turn  gray  if  much  tin  is  present. 
(See  Discussion  22.) 

DISCUSSION 

28.  The    separation    of    arsenic,     antimony    and    tin 
depends  on   the   relative  solubility  of   their  sulphides  in 
HC1.     It  is  necessary,  therefore,   to  follow  the  directions 
very  carefully  in  order  to  secure  a  satisfactory  separation. 
The   solubility   product   of  As2Ss   is   considerably  smaller 
than  that  of  Sb2S5  or  SnS2 ;    hence  it  is  stable  in  a  much 
higher  concentration  of  HC1.      It  is  practically  insoluble 
in  hot  I2N.  HC1  unless  the  H2S  formed  in  the  equilibrium 

As2S5  +  ioHCl^  2(AsCl5)+5H2S 

is  expelled  by  too  vigorous  boiling.  The  hot  acid,  however, 
dissolves  the  tin  very  readily  and  the  antimony  more 
slowly.  In  the  presence  of  much  antimony  some  may 
be  left  undissolved  even  in  the  strong  acid,  but  the  amount 
extracted  will  never  be  so  small  as  to  escape  detection. 
Also,  if  the  acid  becomes  much  diluted,  considerable 
Sb2S5  will  be  left  undissolved.  The  undissolved  portion 
may  be  sufficient  to  give  the  residue  an  orange  color,  but 
it  will  not  interfere  with  the  test  for  arsenic. 

29.  When  both  antimony  and  tin  are  present  and  the 
acid  concentration  is  not  sufficiently  high  or  the  solution 
has  been  allowed  to  cool,  some  tin  may  be  precipitated 
with  the  antimony,  in  which  case  the  color  of  the  H2S 
precipitate  will  usually  be  brown. 

30.  The   magnesia   mixture   used   in   the   detection   of 
arsenic  contains  a  large  excess  of  NH4+  for  the  purpose 
of  reducing  the  OH~  concentration.     This  prevents  the 


PRELIMINARY  EXPERIMENTS  53 

precipitation    of    Mg(OH)2    by  the  NH4OH.     See  Intro- 
duction 13,  also  Appendix  I.) 

31.  Test  lead  is  used  for  the  reduction  of  tin,  rather 
than  zinc  or  iron,  since  lead  does  not  reduce  tin  to  the 
metallic  state.  If  the  stronger  reducing  agent  is  used 
the  tin  will  be  left  in  the  residue  and  will  have  to  be  dis- 
solved in  HC1  before  being  added  to  the  HgCb  solution. 

GROUP  III 

Preliminary  Experiments 

Group  III,  Aluminium  Division 

+  +         +  + 
Al  +  ,  Cr  +  ,  Zn++ 

In  connection  with  the  following  experiments,  study 
Tables  V  and  VI. 

Experiment  17. — Introduce  into  separate  test-tubes 
5-cc.  portions  of  the  test  solutions  containing  the  above 
ions.  Make  each  distinctly  alkaline  with  NH4OH,  then 
saturate  with  H^S.  Heat  nearly  to  boiling  to  coagulate 
the  precipitates.  Write  equations. 

NOTES. — In  the  presence  of  water  A12S3  and  Cr2S3  are  completely 
hydrolized  (see  Introduction  14);  hence  the  hydroxides  formed  on  the 
addition  of  NH4OH  remain  unchanged  when  H2S  is  introduced. 

Zn++  forms  with  NH3  the  complex  ion  Zn(NH3)4++.  (See  Intro- 
duction 15.)  This  removes  enough  Zn++  to  prevent  the  precipitation 
of  Zn(OH)2  but  not  enough  to  prevent  precipitation  of  ZnS  with  HgS. 

Experiment  18. — Decant  off  the  clear  liquid  from  the 
precipitates  of  A1(OH)3,  Cr(OH)3  and  ZnS  (Exp.  17)  and 
dissolve  in  a  slight  excess  of  HC1.  Add  NaOH  slowly  to 
alkaline  reaction,  and  then  about  I  gram  of  solid  Na2C>2, 
a  little  at  a  time.  (Caution:  Do  not  carry  the  Na2O2 
on  paper.  Use  a  dry  watch  glass.  Why?)  Boil  to  expel 
excess  Na2<32.  Write  equations. 


54  CATIONS— (METAL  IONS) 

NOTE. — It  should  be  recalled  (see  note,  Preliminary  Exp.  9,  also 
Introduction  15)  that  a  more  complete  precipitation  does  not  necessarily 
follow  the  addition  of  a  reagent  in  excess  of  that  necessary  to  reach  the 
solubility  product  for  the  compound  concerned.  The  solution  of  A1(OH)3, 
as  well  as  that  of  chromium  and  zinc,  in  an  excess  of  NaOH  presents 
another  example,  but  of  quite  a  different  type.  All  three  belong  to  the 
class  known  as  amphoteric  compounds  (see  Introduction  16).  In  the 
presence  of  excess  NaOH,  therefore,  they  form  the  soluble  sodium  salts 
NaAlO2,  NaCrO2  and  Na2Zn02.  Na2O2  oxidizes  the  NaCrO2  to  the 
more  readily  soluble  Na2CrO4,  but  is  without  effect  on  NaA102  or 
Na2ZnO2.  Why? 

Experiment  19. — Acidify  the  solutions  of  aluminium, 
chromium  and  zinc  (Exp.  18)  with  HNOa  and  add  NH4OH 
to  alkaline  reaction.  Note  the  precipitation  of  Al(OH)s. 
(Difference,  separation  of  Al.)  Write  equations. 

NOTE. — The  action  of  Na2O2  in  Exp.  18  oxidized  the  chromium  to 
the  more  soluble  chromate,  a  compound  which  does  not  possess  ampho- 
teric properties;  hence  on  the  addition  of  HNO3  it  does  not  react  as  a 

base  to  form  a  chromium  salt.     On  the  other  hand,  the  aluminium  and 

+  + 
zinc  return  to  the  positive  radical  as  Al  +    and  Zn++  which  react  with 

the  NH4OH  subsequently  added,  as  outlined  in  Exp.  17. 

Experiment  ipa. — Filter  off  the  precipitate  of  A1(OH)3 
(Exp.  19)  and  wash  with  water.  Dissolve  by  pouring 
5  cc.  of  6N.  HNO3  through  the  filter.  Add  4-5  drops 
of  N.  100  Co(NOs)2  and  evaporate  nearly  to  dryness. 
Soak  up  the  liquid  with  a  small  piece  of  filter  paper.  Roll 
it  up,  wind  a  platinum  wire  around  it  in  the  form  of  a 
spiral  and  heat  in  a  flame  till  all  the  carbon  is  burnt  off. 
A  blue  residue  is  characteristic  of  aluminium. 

NOTE. — The  blue  substance  formed  by  the  interaction  of  Co(N03)2 
and  A1(NO3)3  is  a  compound  of  CoO  and  A12O3,  probably  Co(AlO2)2, 
though  the  exact  composition  has  not  been  determined.  In  carrying 
out  the  reaction  care  must  be  exercised  in  the  addition  of  Co(NO3)2. 
The  aluminium  must  be  in  excess,  otherwise  the  blue  color  is  hidden  by 
the  black  cobalt  oxide.  Furthermore,  the  presence  of  sodium  or  potassium 
salts  causes  the  mass  to  fuse  and  thus  interferes  with  the  test.  They  may 
be  removed  by  washing  the  NH4OH  precipitate  with  water. 


PRELIMINARY  EXPERIMENTS  55 

Experiment  20. — Acidify  the  solution  of  chromium 
and  zinc  (Exp.  19)  with  HC2H3O2,  and  add  BaCU  solu- 
tion. Note  the  precipitation  of  BaCrOi.  (Difference, 
separation  of  Cr.)  Write  equations. 

Experiment  2oa. — Filter  off  the  precipitate  of  BaCrCX 
(Exp.  20)  and  dissolve  it  in  a  very  little  6N.  HNOs.  Dilute 
with  9-10  volumes  of  water  and  to  a  portion  of  it  in  a 
test-tube  add  about  2  cc.  of  ether  and  i  cc.  of  H2O2, 
and  shake. 

NOTE. — The  blue  color  in  the  ether  layer  is  a  perchromic  acid,  probably 
H3Cr07.  It  is  very  unstable,  decomposing  into  oxygen  and  a  chromic 
salt.  Excess  H202  or  acid  accelerates  this  decomposition. 

Experiment  21. — Saturate  the  zinc  solution  (Exp.  20) 
with  H2S.  The  white  flocculent  precipitate  is  ZnS. 
Filter  off  the  precipitate  and  dissolve  it  by  pouring  5  cc. 
of  6N.  HNO3  repeatedly  through  the  filter.  Add  4-5 
drops  of  N.  100  Co(NOs)2  and  evaporate  in  a  porcelain 
dish  almost  to  dryness.  Neutralize  with  Na2COs  solu- 
tion and  add  a  slight  excess.  Evaporate  to  dryness  and 
ignite  gently  till  the  purple  color  of  the  cobalt  disappears. 
Allow  the  residue  to  cool.  The  green  color  is  due  to  a 
compound  of  the  oxides  of  cobalt  and,  zinc,  probably 
CoZnO2. 

NOTE. — ZnS  is  somewhat  more  flocculent  when  precipitated  from  a 
warm  solution.  A  white,  milky-looking  precipitate  does  not  show  the 
presence  of  zinc.  Sulphur  often  separates  out  as  a  white,  m'lky  precipitate, 
especially  on  standing,  or  if  the  current  of  H2S  is  long  continued.  It 
is  often  necessary,  therefore,  to  make  the  confirmatory  test.  The  addition 
of  too  much  Co(NO3)2  must  be  avoided.  (See  note,  Exp. 


56 


CATIONS— (METAL  IONS) 


TABLE  V 

OUTLINE  FOR  THE  SYSTEMATIC  ANALYSIS  OF  GROUP  III 
(SEPARATION  INTO  AL  AND  FE  DIVISIONS) 

Ions  present  in  acid  solution 


No. 

Reagent 

A1+  +  + 

Cr+  +  + 

Zn+  + 

i 

NH4OH 

(A1OH), 

Cr(OH)3 

Zn(NH)4+  + 

2 

H2S 

(A1OH), 

Cr(OH)3 

ZnS 

3 

HC1 

A1C13 

CrCl3 

ZnCl2 

4 

HNO3 

Aldi 

CrCl3 

ZnCl2 

5 

NaOH 

NaAlO2 

NaCrO2 

Na2ZnO2 

6 

Na2O, 

NaAlO2 

Na2CrO4 

Na2ZnO2 

No. 

Mn  +  + 

Fe+  + 

Co+  + 

Ni+  + 

i 

Mn(OH)2 

Fe(OH)2 

Co(NH3)4+  + 

Ni(NH3)4+  + 

2 

MnS 

FeS 

CoS 

NiS 

3 

MnCl2 

FeCl2 

CoS* 

NiS  * 

4 

MnCl2 

FeCl3 

CoCl2 

NiCl2 

5 

Mn(OH)2 

Fe(OH)3 

Co(OH)2 

Ni(OH)2 

6 

MnO(OH)2 

Fe(OH)3 

Co(OH)3 

Ni(OH)2_3 

1 

1 

1 

! 

*  See  Note,  Exp.  23. 


ANALYSIS 
Group  III 

Al  +  ,  Cr  +  ,  Zn++,  Mn++,  Fe++,  Co++,  Ni++ 

Precipitation  and  separation  of  the  aluminium  and  iron 
divisions. 

(50)  Precipitation. — Boil  the  filtrate  from  Group  II 
(20)  till  the  H2S  is  expelled.  Test  the  vapor  with  filter 
paper  wet  with  lead  acetate  [Pb(C2HsO2)2]  solution. 
Add  NH4OH  in  slight  excess,  and  after  shaking  note 
whether  a  precipitate  is  formed.  (See  discussion  32.) 


DISCUSSION    (32-37)  57 

Without  filtering  add  2-3  cc.  more  NH^OH  and  pass 
in  H2S  until,  after  shaking,  the  vapors  blacken  a  piece 
of  filter  paper  moistened  with  Pb(C2HsO2)2  solution. 
Heat  the  mixture  nearly  to  boiling  to  coagulate  the  pre- 
cipitate. Filter  and  wash  the  precipitate  with  water 
containing  about  i  per  cent  of  (NH4)2S.  If  filtration  is 
slow  the  funnel  should  be  kept  covered  with  a  watch  glass 
to  prevent  oxidation  to  soluble  sulphates.  The  filtrate 
should  be  colorless.  (See  Discussion  33.)  Treat  the  pre- 
cipitate by  (51)  and  reserve  the  filtrate  for  analysis  of 
Group  IV  (80). 

(51)  Separation  of  Aluminium  and  Iron  Divisions.— 
Transfer  the  Group  III  precipitate  (50)  with  the  filter  if 
necessary,  to  a  porcelain  dish,  add  5-20  cc.  of  6N.  HC1 
and  stir  for  one  to  two  minutes  in  the  cold.  Heat  the 
mixture  to  boiling  and  if  a  black  residue  still  remains  add 
a  few  drops  of  HNOs  and  boil  again.  Dilute  with  5-10  cc. 
of  water  and  filter  off  the  residue  of  sulphur.  Evaporate 
nearly  to  dryness  to  remove  excess  acid,  dilute  to  10-20  cc., 
and  add  NaOH  to  alkaline  reaction,  avoiding  a  large  excess. 
If  a  very  large  precipitate  forms  add  10—20  cc.  more  water. 
Cool  the  mixture  and  add  1-3  grams  of  solid  Na2O2,  a 
little  at  a  time  and  with  constant  stirring.  (See  Discussion 
35.)  Add  about  5  cc.  of  3N.  Na2CO3  solution  unless 
phosphate  or  the  alkaline  earth  metals  are  known  to  be 
absent.  (See  Discussion  37.)  Boil  to  decompose  excess 
Na2O2;  cool,  dilute  with  an  equal  volume  of  water  and 
filter  with  the  aid  of  suction  if  possible.  (See  Discussion 
36.)  Treat  the  filtrate  by  (60)  and  the  residue  by  (70). 

DISCUSSION 

32.  The  H2S  is  removed  and  the  effect  produced  by 
NH4OH  alone  is  noted,  in  order  to  obtain  information 
regarding  the  presence  of  aluminium  or  other  insoluble 
hydroxides.  NaOH  and  Na2O2,  used  later  in  the  analysis, 
often  contain  small  quantities  of  silica  and  aluminium 


58  CATIONS— (METAL  IONS) 

which  tend  to  make  the  detection  of  aluminium  more  dif- 
ficult. Any  information  gained  'at  this  point,  therefore, 
may  be  of  considerable  importance. 

33.  Ammonium    monosulphide    (  (NH^S)    is    some- 
times used  instead  of  H2S  in  the  precipitation  of  Group  III. 
When  this  is  done  some  NiS  may  be  dissolved  and  pass 
into  the  filtrate,  giving  it  a  brown  or  nearly  black  color. 
The  use  of  H2S  as  directed  in  (50)  almost  completely  pre- 
vents this.     If,  however,  (NH4)2S  is  used  and  the  filtrate 
is  brown  or  nearly  black,  indicating  that  some  NiS  has 
been  dissolved,  it  may  be  precipitated  by  boiling  the  solu- 
tion for  a  few  minutes.     It  should  be  filtered  off  and  added 
to  the  main  precipitate. 

34.  The  presence  of  a  considerable  amount  of  ammo- 
nium salts  not  only  lessens  the  solubility  of  A1(OH)3,  but 
prevents   the    precipitation    of    Mg(OH)2    in    this   group. 
A  sufficient  quantity  of  ammonium  salts  is  usually  formed 
by  the  neutralization  of  the  acid  already  in  the  solution. 
The  presence  of  ammonium  salts  decreases  the  ionization 
of  NH4OH,  owing  to  the  common  ion  effect.     (See  Intro- 
duction  12.)     The  solubility  product  for  the  hydroxides 
of  Group  III  and  also  for  Mg(OH)2  is  so  small  that  even 
a  slight  excess  of  NH4OH  causes  precipitation.     In  the 
presence  of  much  ammonium  salts,  however,  the  concentra- 
tion of  OH~  is   so  reduced  that  its  product  with  that  of 
certain  of  the  metal  ions,  e.g.,  Mn++  or  Mg++,  does  not 
reach  the  solubility  product  for  those  compounds.     In  the 
cases  of  ferric,  aluminium  and  chromium  ions  the  solubility 
product    (CM+++*C3oH~)   is  so  small  that  the  presence  of 
large  amounts  of  ammonium  salts  does  not  appreciably 
affect  their  solubility.     On  the  other  hand,   zinc,   nickel 
and  cobalt  ions  unite  with   NHs   to  form   complex  ions 
similar  to  those  of  silver  and  copper.     (See  note,  Exp.  9 
also  Introduction  15.) 

35.  Na2O2  is  a  very  unstable  substance  and  decomposes 
slowly,  even  in  a  cold  solution,  oxygen  being  given  off. 
In  a  hot  solution  the  decomposition  may  become  explosive. 


DISCUSSION  (32-37)  59 

The  peroxide  should  therefore  be  added  in  very  small 
quantities  to  a  cold  solution.  This  may  be  easily  done 
by  transferring  a  little  of  it  to  a  dry  watch  glass  and 
sprinkling  it  into  the  solution,  stirring  constantly.  The 
reaction  is  known  to  be  complete  when  a  steady  stream 
of  gas  is  evolved  after  the  mixture  has  been  thoroughly 
stirred.  If  much  chromium  is  present  the  Na2O2  should 
be  added  until  the  green  chromic  salt  has  been  entirely 
changed  to  the  yellow  chromate.  The  solution  should 
be  diluted  before  filtering  in  order  to  prevent  disinte- 
gration of  the  filter  paper  by  the  strong  alkali. 

36.  The  separation  of  the  aluminium  and  iron  divisions 
by  means  of  NaOH,  Na2O2  and  Na2CO3  is  very  satisfactory 
except  in  the  case  of  zinc.     When  much  of  the  iron  division 
is  present  15—20  mgs.  of  zinc  may  be  carried  down  in  the 
precipitate,  so  that  provision  must  be  made  in  the  iron 
division  for  its  detection. 

37.  It  should  be  remembered  that  the  phosphates  of 
barium,  strontium,  calcium  and  magnesium  are  insoluble 
in   alkaline   solution,    and    hence   may   have   precipitated 
along  with  Group   III,  in  case  phosphate  was  present  in 
the  original  material.       Na2CO,3  is  used  as  a  reagent  in  the 
separation  of  the  aluminium  and  iron  divisions   to  insure 
complete  precipitation  of  these  elements  since  their  hydrox- 
ides are  somewhat  soluble  even  in  strong  NaOH  solution. 
ZnCOs,   though  insoluble  in  dilute  solutions  of  Na2COs 
alone,  is  soluble  in  the  presence  of  NaOH,  owing  to  the 
formation    of    the    zincate    ion    (ZnO2~).     Na2CO3    also 
serves  to  decompose  the  chromates  of  the  alkaline  earths 
and  so  prevent  the  precipitation  of  chromium. 


60 


CATIONS— (METAL  IONS) 


TABLE  VI( 

OUTLINE  OF  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  III 

(Al  DIVISION) 


No. 

Reagent 

NaAlO, 

Na2CrO4 

Na£ZnO2 

i 

2 

3 

4 

5 

HNOi 

NH4OH 
HNO3 
Co(N03)2 
HC2H3O2 
BaCl2 
HNO3 
Ether 
H,02 
H2S 
HNO3 
Co(N03)2 
Na2C03 

A1(N03)3 
M(OH), 

Co(A102)2(?) 

Na2Cr2O7 
Na2CrO4 

Na2Cr2O7 
BaCrO4 

Zn(N03)2 

Zn(NH3)4(N03)2 
1 

Zn(N03)2 
ZnCl2 

I 

ZnS 
Zn(NO,), 

CoZnO2(?) 

1 

H2Cr04 
H3Cr07(?) 

1 

ANALYSIS 

Group  III,  Aluminium  Division 

(60)  Separation    of   Aluminium. — Acidify    the    nitrate 
from  (51)  with  i6N.  HNO3  avoiding  a  large  excess;    add 
NH-iOH  just  to  alkaline  reaction  and  then  2-3  cc.  more. 
(See  Discussion  38.)     The  formation  of  a  white  flocculent 
precipitate  shows  the  presence  of  aluminium.     (See  Dis- 
cussion 39.)     Heat  the  mixture  almost  to  boiling  to  coagu- 
late the  precipitate,  filter  and  wash  thoroughly  with  hot 
water.     Unless    the    precipitate    is    white    and    flocculent 
the  confirmatory^fcifct  should  be  made.     Treat  the  filtrate 
by  (62). 

(61)  Confirmatory  Test  for  Aluminium. — Dissolve  the 
precipitate  (60)  in  5  cc.  of  6N.  HNO3 ;    add  4-5  drops  of 
N.  100  Co(NO3)2   or   less   if   the   precipitate   was   small, 
evaporate  almost  to  dryness  and  add   1-2  cc.  of  water. 
Absorb  the  solution  in  a  small  piece  of  filter  paper,  wind 


OUTLINE  OF  ANALYSIS  (TABLE  VI)  61 

a  platinum  wire  around  it  in  the  form  of  a  spiral  and  ignite 
the  paper  in  a  flame  until  the  carbon  is  completely  burned. 
A  blue  residue  shows  the  presence  of  aluminium.  (See 
note,  Exp.  iga.) 

(62)  Detection   of   Chromium. — To   the   filtrate   form 
(66)  add  HC2H3O2  slowly  until  the  solution  is  just  acid. 
(Test  with  litmus.)     If  the  solution  is  colorless,  chromium 
is  absent  and  the  solution  should  be  tested  for  zinc  accord- 
ing to  (64).     If  the  solution  is  at  all  yellow,  add  about 
IO  cc.  of  BaCl2  solution  and  heat  the  mixture  to  boiling. 
If  a  yellow  precipitate  forms,  chromium  is  present.     Unless 
the  precipitate  is  distinctly  yellow  the  confirmatory  test 
should  be  made.     (See  Discussion  40.)     Treat  the  filtrate 
by  (64). 

(63)  Confirmatory  Test  for  Chromium. — Dissolve  the 
precipitate  (62)  by  pouring  repeatedly  through  the  filter 
a  cold   lo-cc.   portion  of  o.6N.  HNOs.     To  the  solution 
contained  in  a  test-tube  add  1-2  cc.  of  ether  and  i  cc.  of 
3  per  cent  H2O2  solution.     Shake  the  mixture.     A  blue 
color  appearing  in  the  ether  layer  shows  the  presence  of 
chromium.     (See  note,  Exp.  2Oa.) 

(64)  Detection  of  Zinc. — Warm  the  HC2H3O2  solution 
(62)  or  the  filtrate  from  (62)  to  about  50°  and  saturate  in 
a  small  flask  with  H2S.     If  a  flocculent  precipitate  does 
not  form  at  once,  cork  the  flask  and  allow  it  to  stand 
five  to  ten  minutes.     The  formation  of  a  white  flocculent 
precipitate  shows  the  presence  of  zinc.     Unless  the  pre- 
cipitate   is   white    and    flocculent   the    confirmatory    test 
should  be  made. 

(65)  Confirmatory  Test  for  Zinc. — Filter  off  the  pre- 
cipitate obtained  in   (64)  and  dissolve  it  by  pouring  re- 
peatedly through   the  filter  a  5-cc.  portion  of  6N.  HNO3. 
To  the  resulting  solution  add  4-5  drops  of  N.  100  Co(NO3)2, 
or  less  if  the  precipitate  was  small.     Evaporate  the  mix- 
ture  almost    to   dryness   in   a   porcelain   dish,    neutralize 
with  Na2COa  solution  and  add  0.5  cc.  in  excess.    -Evapo- 
rate to  dryness  and  ignite  gently  until  the  purple  color 


62  CATIONS— (METAL  IONS) 

due  to  the  Co(NO3)2  disappears.  (See  Discussion  41.) 
The  appearance  of  a  green  color  shows  the  presence  of 
zinc. 

DISCUSSION 

38.  In  the  separation  of  aluminium  a  large  excess  of 
NH4OH  must  be  avoided,  since  it  tends  to  redissolve  the 
A1(OH)3  by  forming  NH4AlOo.     A  moderate  excess,  how- 
ever, must  be  present  in  order  to  keep  the  zinc  in  solution. 

39.  Since  aluminium  and  silica  are  often  present  in  the 
Na2O2  and  NaOH  used  as  reagents  in  the  separation  of 
the  aluminium  and  iron  divisions,  a  blank  test  should  be 
made  for  these  impurities  by  treating  10-15  cc-  of  water, 
to  which  has  been  added  about  the  same  quantities  of 
these  two  reagents  as  was  used  in  the  regular  analysis, 
by  (60)  and  comparing   the   precipitate  formed  with  that 
obtained  in  the  regular  analysis.     The  confirmatory  test 
should  always  be  made  in  case  the  NH4OH  precipitate  is 
small,    in  order  to  avoid  mistaking  silicic  acid    (H^SiOs) 
for  A1(OH)3. 

40.  If,  owing  to  careless  manipulation  in  filtering  and 
washing  the  second  or  third  group  precipitates,  and  sul- 
phides have  been  oxidized  to  sulphates,   the  addition  of 
BaCb  for  the  separation  of  chromium  may  give  a  white 
or   pale   yellow   precipitate   composed   largely   of    BaSO4. 
This  not  only  makes  it  necessary  to  confirm  the  presence 
of  chromium  but  renders  the  confirmatory  test  less  delicate 
owing  to  the  difficulty  of  extracting  the  barium  chromate 
(BaCrO4)  from  the  BaSO4  with  acid.     The  color  of  the 
solution  after  treatment  with  NaOH  and  Na2C>2  for  the 
separation    of   aluminium    and    iron   divisions    should    be 
noted.     A  yellow  color,  changing  to  orange  on  the  addition 
of  acid,  indicates  chromium. 

41.  The   confirmatory  test  for  zinc  is  of  value   only 
when  the  IrbS  precipitate  is  small  and  finely  divided  or 
when  the  presence  of  foreign  materials  causes  it  to  be 


PRELIMINARY  EXPERIMENTS  63 

dark-colored.  In  the  hands  of  the  careful  worker  the  test 
is  quite  satisfactory  and  will  detect  0.5  mg.  of  zinc.  If, 
owing  to  too  much  heat,  the  residue  becomes  black,  it 
should  be  dissolved  in  a  few  drops  of  HNOa  and  evaporated 
almost  to  dryness,  and  the  test  should  be  repeated  with 
the  exception  that  no  more  Co(NO3)2  should  be  added. 

Prelminary  Experiments 

Group  III,  Iron  Division 

Mn++,  Fe++,  Co++ 


In  connection  with  the  following  experiments  study 
Tables  V  and  VII. 

Experiment  22.  —  To  separate  5-cc.  portions  of  the  test 
solutions  containing  the  above  ions,  add  NH4OH  to  alka- 
line reaction  and  saturate  with  FhS.  Note  the  color  of 
the  precipitates  formed.  Write  equations. 

Experiment  23.  —  Decant  or  filter  off  the  liquid  from 
the  precipitates  (Exp.  22)  and  dissolve  by  the  addition 
of  HC1.  Add  a  few  drops  of  HNO3  if  HC1  fails  to  effect 
a  solution.  Write  equations. 

NOTE.  —  CoS  and  NiS  are  dissolved  very  slowly  by  HCl  but  much 
more  rapidly  in  aqua  regia.  The  oxidizing  action  of  aqua  regia  removes 
the  S=,  thus  shifting  the  equilibrium  and  allowing  the  cobalt  and  nickel 
to  pass  into  solution. 

Experiment  24.  —  To  the  clear  solutions  (Exp.  23) 
add  NaOH  to  alkaline  reaction  and  then  about  I  gram 
of  solid  Na2C>2,  a  little  at  a  time.  Note  any  changes 
in  the  precipitates  caused  by  the  Na2C>2.  Boil  to  expel 
excess  Na2C>2.  Write  equations. 

NOTE.—  Na202  oxidizes  Mn(OH)2  to  the  less  soluble  MnO(OH)2, 
Fe(OH)2  to  Fe(OH)3  and  Co(OH)2  to  Co(OH)3.  Ni(OH)2  is  only  parti- 
ally oxidized  to  Ni(OH)3.  All  these  compounds  are  insoluble  in  an  excess 
of  NaOH,  differing  in  this  respect  from  the  hydroxides  of  the  aluminium 
division. 


64  CATIONS— (METAL  IONS) 

Experiment  25. — Decant  or  filter  off  the  liquid  from 
the  precipitates  (Exp.  24)  and  add  5-10  cc.  of  HNO3.  Note 
the  fact  that  iron,  cobalt  and  nickel  hydroxides  readily 
dissolve.  Heat  the  mixture  containing  the  manganese 
nearly  to  boiling  and  add  H^Oo,  a  few  drops  at  a  time, 
until  solution  is  complete.  Write  equations. 

NOTE. — MnO(OH)2  is  not  affected  by  HN03  alone,  but  if  reduced 
to  a  lower  oxide  it  is  dissolved  as  Mn(N03)2.  The  reduction  is  readily 
brought  about  with  H202,  oxygen  being  evolved. 

Experiment  26. — Heat  the  solutions  (Exp.  25)  to  boil- 
ing and  add  gradually  1-2  grams  of  solid  KC1O3  or  NaClOa. 
Boil  for  a  minute  or  two.  Note  the  precipitation  of  MnO2. 
(Difference,  separation  of  Mn.)  Write  equations. 

Experment  26a. — Filter  off  the  liquid  from  the  MnC^ 
(Exp.  26),  add  5-10  cc.  of  i6N.  HNOs  and  1-2  grams  of 
Boil  for  a  minute  or  two  and  set  aside  till  the 
settles.  The  purple  color  of  the  solution  is  due 
to  HMnO4.  Write  equations. 

Experiment  27. — To  the  solutions  of  iron  cobalt  and 
nickel  (Exp.  26)  slowly  add  NH4OH  to  distinctly  alkaline 
reaction.  Note  the  precipitation  of  Fe(OH)s.  (Differ- 
ence, separation  of  Fe.) 

NOTE. — The  Co(OH)2  and  Ni(OH)2  at  first  formed  dissolve  in  an  excess 
of  NH4OH  with  the  formation  of  the  complex  ions  Co(NH3)4+  +  and 
Ni(NH3)4++.  (See  Introduction  15.) 

Experiment  2ya. — Decant  or  filter  the  liquid  from  the 
Fe(OH)a  precipitate  (Exp.  27)  and  dissolve  it  in  a  small 
quantity  of  6N.  HC1.  Divide  the  solution  into  two  parts 
and  add  K4Fe(CN)6  to  the  one  and  KCNS  to  the  other. 
Write  equations. 

NOTE.— It  is  important  to  use  HC1  as  the  solvent  for  Fe(OH)3-  HN03 
must  not  be  used  since  the  subsequent  reaction  with  KCNS  is  rendered 
much  less  delicate  in  its  presence,  N02,  a  common  impurity  in  HNO3, 
also  giving  a  red  color  with  KCNS.  The  test  is  extremely  delicate,  and 


PRELIMINARY  EXPERIMENTS  65 

when  only  a  faint  red  color  is  produced  the  acids  used  should  be  tested 
for  iron.  The  red  color  is  due  to  the  formation  of  non-ionized  ferric 

thiocyanate  [Fe(CNS)3]  and  the  test  is  therefore  rendered  more  delicate 

+  + 

when  an  excess  of  the  reagent  is   used.     K4Fe(CN)6   gives  with   Fe  + 
a  deep-blue  precipitate  of  Fe4(Fe(CN)6)3  (Prussian  blue). 

Experiment  28. — Saturate  the  solutions  of  cobalt  and 
nickel  (Exp.  27)  with  H2S.  Decant  or  filter  the  liquid 
from  the  precipitates  of  CoS  and  NiS  and  dissolve  by  the 
addition  of  HC1  and  a  few  drops  of  HNOa.  Evaporate 
to  1-2  cc.  and  just  neutralize  with  NaOH.  Add  HC2H3O2 
to  very  faintly  acid  reaction,  and  a  considerable  excess  of 
potassium  nitrite  (KNp2).  Note  the  precipitation  of 
potassium  cobaltinitrite  (K3Co(NO2)6).  (Difference,  detec- 
tion of  Co.) 

NOTE. — In  very  faintly  acid  reaction,  cobalt  produces  with  KN02  a 
yellow  crystalline  precipitate  of  K3Co(NO2)6;  hence  the  necessity  of 
neutralizing  the  strong  acid  with  NaOH  and  subsequently  rendering 
the  solution  acid  with  the  weak  HC2H3O2.  The  precipitate  may  form 
rather  slowly.  The  reaction  preceeds  as  follows: 

Co(NO3)2+7KNO2+2HC2H3O2->K3Co(NO2)6-i-NO+2KC2H302+2KN03 

A  part  of  the  KN02  is  used  in  oxidizing  the  cobalt  to  the  trivalent  con- 
dition. This  then  unites  with  the  excess  KN02  to  form  the  complex 
salt  K3Co(N02)6.  The  K3Co(N02)6  is  somewhat  soluble  in  water  but 
much  less  so  in  KN02  solution,  owing  to  the  common  ion  effect  of  K+. 

Experiment  29. — To  the  solution  containing  nickel 
(Exp.  28)  add  NH4OH  to  faintly  alkaline  reaction,  heat 
to  boiling  to  drive  off  excess  NHs  and  then  add  a  few  drops 
of  dimethyl-glyoxime.  Note  the  precipitation  of  nickel 
dimethyl-glyoxime  [NiKCHs^CiN^Hh].  (Difference, 
detection  of  Ni.) 


66 


CATIONS— (METAL  IONS) 


TABLE  VII 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  III 

(Fe  DIVISION) 


No. 

Reagent 

MnO(OH)2 

Fe(OH)3 

Co(OH)3 

Ni(OH)2 

2 

3 

4 

5 
6 

7 

8 

HNO3 

H202 
KC103 
HN03 
Pb02 
*NH4OH 
HC1 
(o)    K4Fe(CN)6 
(b)    KCNS 
H2S 
HC1—  KC1O3 
NaOH 
HC2H302 
KNO2 
NH4OH 
(CH3)2C2(NOH)2 

MnO(OH)2 
Mn(NO3)2 
MnO-2 

Fe(N03)3 
Fe(N03)3 
Fe(N03)3 

Fe(OH)3 
FeCl3 

Fe4(Fe(CN)6)3 

Co(N03)2 
Co(N03)2 
Co(N03)2 

Co(NH3)4(NO3)2 

J 

CoCl2 
KsCo(N02)6 

I 

Ni(N03)2 
Ni(N03)2 
Ni(N03)2 

Ni(NH3)4(N03)2 

NiS 
NiCl2 

K4Ni(N02)6 
Ni(CH3)2C2N2O2H)2 

HMnO4 

Fe(CNS)3 

*  If  phosphates  may  be  present,  the  procedure  should  follow  Table  VIII. 


TABLE  VIII 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  Fe,  Co  AND  Ni 
WHEN  PHOSPHATES  MAY  BE  PRESENT 

Ions  present  in  HNO3  solution 


Reagent 

po4— 

Fe+++ 

Co++ 

NI++ 

i   (a)   (NH4)2MoO4 
(b)   HC1 
+  K4Fe(CN)6 
(c)    (NH4OH 
+  HC2H302 
(NH4C2H302 
+  FeCl3 
2  NH4C2H3O2 

3  H2S 

(NH4)3P04(Mo03),2 

Fe+++ 

Fe4(Fe(CN)6)3 

Fe(C2H302)3 
Fe(OH)2C2H302 

Co++ 
Co++ 

Co++ 

Co(C2H302)2(?) 

CoS 

Ni++ 

Ni++ 

Ni++ 
Ni(C2H302)2(?) 

NiS 

FeP04 

i 

i 

OUTLINES  OF  ANALYSIS  (TABLES  VII,  VIII)  67 

ANALYSIS 

Group  III,  Iron  Division 

(70)  Precipitation  of  Manganese. — Transfer  the  pre- 
cipitate obtained  in  (51)  to  a  porcelain  dish,  together  with 
the  filter  if  necessary;    add  5-20  cc.  of  6N.  HNOs,  heat 
nearly  to  boiling  and,  if  any  of  the  precipitate  remains 
undissolved,  add  slowly  and  with  constant  stirring  3  per 
cent  H2O2  solution  until  the  precipitate  has  completely 
dissolved.     Filter    to    remove    the    paper    and    evaporate 
almost  to  dryness.     Add   10-15  cc.  of  i6N.  HNO3,  heat 
to  boiling,  add  0.5-1  gram  of  powdered  KC1O3  and  boil 
gently.       If    a    large    precipitate    forms    add    1-2    grams 
more  KClOs,  a  small  portion  at  a  time.     A  dark  brown 
or    black   precipitate   shows    the   presence  of  manganese. 
Filter  through  an  asbestos  filter  (see  Discussion  42).     Heat 
the  filtrate  to  boiling  and  add  0.5  gram  more  of  powdered 
KC1O3.     Boil    gently.     If   a    precipitate    forms    add    1-2 
grams  more   KClOs,   heat  to   boiling  and   filter  through 
the  same  filter.     Wash  the  precipitate  with  a  little  i6N. 
HNOa  which  has  been  freed  from  the  oxides  of  nitrogen 
by  warming  with  a  little  KClOs.     Treat  the  precipitate 
by  (71)  and  the  filtrate  by  (72). 

(71)  Confirmatory  Test  for  Manganese. — Transfer  the 
precipitate   (70)   to  an  evaporating  dish,  add   1-2  grams 
of  lead  dioxide  (PbO2)  and  10-15  cc.  of  6N.  HNO3.     Heat 
the  mixture  to  boiling  and  boil  for  one  to  two  minutes. 
Then  pour  into  a  test-tube  and  allow  the  PbO2  to  settle. 
A  purple  solution  shows  manganese  to  be  present 

(72)  Detection     of     Phosphate.  —  Heat     about    one- 
tenth  of  the  filtrate  from  (70)  to  boiling  and  pour  it  into 
about  three  times   its  volume  of  ammonium   molybdate 
[(NH4)2MoO4]  reagent.   (See  Discussion  43.)  The  formation 
of  a  fine  yellow  precipitate  shows  the  presence  of  phosphate. 

If  phosphate  is  shown  to  be  present  treat  a  second 
one- tenth  portion  of  the  filtrate  by  (75). 


68  CATIONS— (METAL  IONS) 

If  phosphate  is  absent  treat  the  remaining  nine- tenths 

by  (73). 

(73)  Separation  of  Iron  in  the  Absence  of  Phosphate.— 
If  phosphate  was  shown  to  be  absent  add  NH.tOH  to  the 
remaining  nine-tenths  of  filtrate  (70)  until  it  is  distinctly 
alkaline   and    then   4-5   cc.  in  excess.     The    formation  of 
a  reddish-brown  precipitate  shows  the  presence  of  iron. 
Filter   and   wash   the   precipitate.     Treat   the   filtrate   by 
(77)  and  the  precipitate  by  (74). 

(74)  Confirmatory  Test  for  Iron. — Dissolve   the   pre- 
cipitate  (73)   by  pouring  a  5-10  cc.   portion  of  6N.  HC1 
repeatedly  through  the  filter.     To  the  resulting  solution 
add  a  few  drops  of  potassium  ferrocyanide  [K4Fe(CN)6] 
solution.        A     deep-blue      precipitate     (Prussian      blue, 
Fe4(Fe(CN)6)3)  shows  the  presence  of  iron. 

(75)  Detection  of  Iron  in  the  Presence  of  Phosphate.— 
If  phosphate  was  shown  to  be  present  evaporate  about 
one-tenth  of  the  filtrate  (70)  to  dryness,  add   1-2  cc.  of 
I2N.  HC1  and  evaporate  to  dryness  again.     Dissolve  the 
residue  in  5-6  cc.  of  6N.  HC1  and  add  5  cc.  of  potassium 
thiocyanate    (KCNS)    solution.     A   red   coloration   shows 
the    presence    of    iron.     Treat    the    remainder   of    filtrate 
(70)  by  (76). 

(76)  Removal    of   Phosphate. — To    the    remainder   of 
the  filtrate  from   (70)  add  NH4OH  slowly  and  with  fre- 
quent stirring,   until  the  precipitate  formed  just  fails  to 
redissolve.     The    solution    should    not    be    alkaline.     If, 
owing  to  the  addition  of  too  much  NH4OH,  it  has  become 
alkaline  or  a  large  precipitate  separates,  H^HaCb  should 
be  added  to  distinct  acid  reaction.     Add   15  cc.  of  3N. 
NH4C2H3O2    solution    and    unless    the    mixture    has    a 
brownish-red    color   add    ferric   chloride    (FeCls)    solution 
drop   by  drop,    until   a   brownish-red   color   is   produced. 
(See  Discussion  44.)     Transfer  the  mixture  to  a  250  cc. 
flask,  add  water  to  make  a  total  volume  of  about  100  cc. 
and  boil  for  five  minutes,  adding  more  water  if  a  large 
precipitate   separates.     Allow  the   mixture  to   stand   one 


OUTLINES  OF  ANALYSIS  (TABLES  VII,  VIII)  69 

to  two  minutes,  filter  while  still  hot  and  wash  with  hot 
water.  To  the  filtrate  add  10  cc.  of  3N.  NH^H.sCb 
solution  and  boil.  If  a  precipitate  separates  filter  it  off 
through  a  separate  filter  and  reject.  Make  the  filtrate 
alkaline  with  NH4OH  and  treat  it  by  (77). 

(77)  Separation    of   Zinc,    Cobalt   and    Nickel. — Pass 
H2S  into  the  filtrate  (73)  or  (76)  until,  after  shaking,  the 
vapors  blacken  filter  paper  moistened  with  lead  acetate 
[Pb(C2H3O2)2]  solution.     The  formation  of  a  black  pre- 
cipitate indicates  the  presence  of  nickel  or  cobalt  or  both; 
a  white  flocculent  precipitate  indicates  zinc.     Filter,  wash 
the    precipitate  with  water    containing    a    few  drops  of 
(NH4)2S  and  treat  it  by   (77a)   unless  zinc  has  already 
been  found;   otherwise  treat  it  by  (78).     Treat  the  filtrate 
by  (80)  if  phosphate  or  much  chromium  has  been  found. 
Otherwise  reject  it.     (See  Discussion  45.) 

(77a)  Separation  of  Zinc  from  Cobalt  and  Nickel. — If 
zinc  has  already  been  found  this  procedure  may  be  omitted 
and  the  precipitate  from  (77)  treated  directly  by  (78). 
If  zinc  has  not  already  been  found,  transfer  the  precipitate 
(77),  with  the  filter  if  necessary,  to  an  evaporating  dish, 
add  10-30  cc.  of  iN.  HC1  and  stir  the  cold  mixture  for 
about  five  minutes.  Filter,  wash  the  residue  and  treat 
it  by  (78).  Boil  the  filtrate  until  the  H2S  is  completely 
expelled.  Make  the  solution  alkaline  with  NaOH.  To 
the  cold  solution  add,  with  constant  stirring,  about  I  cc. 
of  powdered  Na2O2,  a  little  at  a  time.  Boil  to  decompose 
excess  Na2C>2,  cool  and  filter.  Unite  the  residue  with 
that  undissolved  by  the  iN.  HC1  above  and  treat  by  (78). 
Acidify  the  filtrate  with  HC2H3O2  and  test  it  for  zinc 
according  to  (64).  (See  Discussion  46.) 

(78)  Detection  of  Cobalt. — Transfer  the  combined  resi- 
dues from  (77a)  to  a  porcelain  dish  and  add  5-15  cc.  of 
6N.  HC1.     Heat  the  mixture  nearly  to  boiling,  and  while 
hot  sprinkle  into  it  a  little  powdered  KC1O3.     When  the 
reaction  is  complete  and  only  a  residue  of  sulphur  remains, 
filter  and  evaporate  the  filtrate  almost  to  dryness.     Take 


70  CATIONS— (METAL  IONS) 

up  the  residue  with  1-2  cc.  of  water  and  make  the  solution 
just  neutral  with  NaOH.  (Test  with  litmus.)  Add  2-3 
cc.  of  6N.  HC2H3O2  and  then  3-5  cc.  of  potassium  nitrite 
(KNO2)  solution  and  let  the  mixture  stand,  with  occasional 
shaking,  for  fifteen  to  twenty  minutes.  (See  Discussion 
47.)  The  formation  of  a  yellow  crystalline  precipitate 
shows  the  presence  of  cobalt.  Filter  and  treat  the  nitrate 
by  (79). 

(79)  Detection  of  Nickel. — Dilute  the  filtrate  from  (78) 
to  about  25  cc.,  add  NH4OH  until  the  solution  is  just 
alkaline.  Heat  to  boiling  to  expel  excess  NH3  and  then 
add  a  few  drops  of  dimethyl-glyoxime  [(CH3CNOH)2] 
solution,  and  if  a  red  precipitate  or  coloration  does  not 
form  at  once  allow  the  mixture  to  stand  five  to  ten  minutes. 
A  red  precipitate  shows  the  presence  of  nickel.  (See 
Discussion  48.) 

DISCUSSION 

42.  Owing  to   the   oxidizing  action   of   strong   HNO3, 
the  precipitate  of   MnO2   cannot   be   filtered   through  an 
ordinary  filter  paper.     An  asbestos  filter  may  be  prepared 
by  placing  a  small  pinch  of  glass  wool  in  the  neck  of  a 
funnel,    tamping   it   gently   with   the   finger   and   pouring 
over  it  a  suspension  of  fine  asbestos  fiber  in  water,  enough 
to  make  a  layer  2-3  mm.  thick. 

43.  If  the  solution  containing  the  ammonium  molyb- 
date  is  heated  much  above  60°  C.,  white  insoluble  molybdic 
acid   (MoO3)   may  separate  out  and  render  the  test  for 
phosphate    much    less    delicate.     The    yellow    precipitate 
is    (NH4)3PO4-i2MoO3    and    is    most   readily   formed    at 
about  60°  in  the  presence  of  a  large  excess  of  (NH4)2MoO4 
and  HNO3. 

44.  The  separation  of  ferric  iron  and  phosphate  from 
the   bivalent   elements   is   based   on   the   following   facts: 
The  solubility  product  for  FePO4  is  much  smaller  than  that 


DISCUSSION  (42-48)  71 

for  the  phosphates  of  the  bivalent  elements.  In  a  boiling 
acetic  acid  solution  containing  a  large  excess  of  the  acetate 
ion,  insoluble  basic  ferric  acetate  [Fe(OH)2C2H3O2]  is 
formed.  In  order  to  issure  complete  removal  of  the  phos- 
phate and  prevent  its  interaction  with  the  bivalent  elements 

-f + 

excess  Fe  +  is  necessary.  This  is  shown  by  the  brownish- 
red  color  of  Fe(C2HsO2)3.  The  separation  is  satisfactory 
and  complete  if  accurately  carried  out.  If  the  solution 
is  allowed  to  become  alkaline  much  of  the  iron  will  pre- 
cipitate ^s  Fe(OH)3  while  the  bivalent  elements  may  form 
insoluble  phosphates  or  hydroxides.  On  the  other  hand, 
if  the  solution  is  too  acid  the  FePO4  will  not  be  precipi- 
tated owing  to  the  formation  of  HPO4=  and  the  consequent 

+  + 

reduction  of  the  ion  product  CFe  +  -CP0^.  If  there  is 
a  tendency  toward  the  formation  of  colloidal  FePCU  its 
coagulation  may  be  greatly  promoted  by  boiling. 

45.  The  solution  obtained  after  the  removal  of  zinc, 
cobalt  and  nickel  may  contain  some  or  even  all  of  the 
alkaline    earth    metals.     In    the    presence    of    phosphate, 
barium,  strontium,  calcium  and  magnesium  may  be  pre- 
cipitated as  phosphates  along  with  the  metals  of  Group  III. 
In  the  presence  of  much  chromium,  magnesium  may  be 
precipitated  completely  as  Mg(CrO2)2. 

46.  Although  cobalt  and  nickel  sulphides  are  soluble 
in  HC1  the  reaction  is  comparatively  slow  even  in  moder- 
ately concentrated  solutions.     Therefore,  when  a  mixture 
of  the  sulphides  of  zinc,  cobalt  and  nickel  is  treated  with 
a  cold  iN.  solution  of  HC1  the  greater  portion  (80-90  per 
cent)  of  the  cobalt  and  nickel  remains  undissolved,  while 
all    of    the    zinc    passes    into    solution.     The    subsequent 
treatment  of  the  solution  with  NaOH  and  Na2O2  gives 
a  very  satisfactory  separation  since  the  cobalt  and  nickel 
are  present  in  such  small  quantities   that   only  an  insig- 
nificant amount  of  zinc  is  carried  down  with  them. 

47.  It  has  been  mentioned   (see  note,   Exp.  28)   that 


72  CATIONS— (METAL  IONS) 

when  KNO2  is  added  to  an  acid  solution  containing  cobalt 

and  nickel  a  part  of  the  HNC>2  formed  oxidizes  the  cobalt 

+  + 
from  the  cobaltous  (Co4"4")  to  the  cobaltic    (Co  +  )    state. 

The  cobaltic  ion  in  turn  unites  with  the  nitrite  ion  to  form 

+  + 

the  complex  ion  Co(NO2)e  +  •  In  the  presence  of 
HC2H3O2  the  solubility  product  for  K3Co(NO2)6  is  soon 
reached,  resulting  in  the  precipitation  of  potassium  cobalt- 

initrite  [K3Co(NO2)6].     Nickel  is  not  readily  oxidized  by 

+  + 
HNO2  to  the  nickelic  (Ni  +  )  state  although  it  does  form 

the  complex  ion  Ni(NO2)e++.  The  potassium  salt 
K4Ni(NO2)e  is  fairly  soluble,  however,  so  that  little  or  no 
nickel  is  precipitated  with  the  cobalt.  Since  K3Co(NO2)6 
is  readily  soluble  in  strong  acids  the  necessity  of  neutral- 
izing the  HC1  and  subsequently  acidifying  with  the 
weaker  HC2H3O2  is  apparent  and  since  both  cobalt  and 
nickel  form  complex  ions  with  NC>2~  a  large  excess  of 
KNO2  must  be  added. 

48.  Dimethyl-glyoxime  is  a  weak  monobasic  organic 
acid  having  the  formula  (CH3)2C2(NOH)2.  It  reacts 
with  the  nickel  ion  (Ni  +  +)  according  to  the  following 
equation : 

CH3— C— NOH         CH3— C— NOH       HON— C— CH3 
Ni+++2  :  :  :  +2H+ 

CH3— C— NOH        CH3— C— NO— Ni— ON— C— CH3 

Nickel  dimethyl-glyoxime  is  least  soluble  in  a  neutral  or 
weakly  acid  solution,  is  very  voluminous  and  has  an  intense 
red  color.  It  is  capable,  therefore,  of  detecting  very 
small  quantities  (o.i  mg.)  of  nickel. 

In  the  presence  of  cobalt,  dimethyl-glyoxime  produces 
a  brown  coloration  which  deepens  as  the  amount  of  cobalt 
is  increased.  It  is  advisable,  therefore,  to  remove  the 
cobalt  before  making  the  test  for  nickel. 


PRELIMINARY  EXPERIMENTS  73 

GROUP  IV 

Preliminary  Experiments 
Ba++,  Sr++,  Ca+  + 

In  connection  with  the  following  experiments  study 
Table  IX. 

Experiment  30. — To  separate  5-cc.  portions  of  the 
test  solutions  containing  the  above  ions  add  NH4OH, 
drop  by  drop,  till  its  odor  persists  after  shaking.  Heat 
to  boiling  and  then  add  ammonium  carbonate  [(NH4)2CO3] 
in  slight  excess.  Allow  the  precipitates  to  settle,  decant 
off  the  liquid  and  dissolve  the  precipitates  in  HC2H3O2. 
Write  equations. 

Experiment  31. — Divide  the  solutions  (Exp.  30)  into 
two  portions  each.  To  one  add  an  equal  volume  of  satu- 
rated calcium  sulphate  (CaSO4)  solution.  If  a  precipitate 
does  not  form  immediately  heat  to  boiling.  Note  the 
character  of  the  precipitates  and  the  relative  time  for 
precipitation.  Write  equations. 

Experiment  32. — Determine  the  color  of  the  flame 
produced  by  barium,  strontium  and  calcium  by  introducing 
a  little  of  each  (second  portion,  Exp.  31)  on  a  clean  platinum 
wire  into  the  colorless  flame  of  the  Bunsen  burner.  (Cau- 
tion: the  platinum  wire  should  be  held  only  in  the  outer 
cone  of  the  Bunsen  flame.) 

Experiment  33. — To  the  acetic  acid  solutions  of  the 
above  metals  (second  portion,  Exp.  31)  diluted  to  about 
50  cc.  add  potassium  dichromate  (K^C^O?)  solution. 
Note  the  precipitation  of  BaCrO4.  (Difference,  separation 
of  Ba.)  Write  equations. 

NOTE. — The  solubility  of  the  chromates  of  barium,  strontium  and 
calcium  in  100  cc.  of  water  at  10°  is  0.38,  120  and  400  mgs.  respectively 
From  this  it  is  evident  that  in  comparatively  dilute  solutions  practically 
no  strontium  or  calcium  will  be  precipitated.  From  more  concentrated 
solutions,  however,  some  strontium  may  separate  out  as  SrCr04. 


74 


CATIONS— (METAL  IONS) 


Experiment  34. — To  each  of  the  solutions  containing 
strontium  and  calcium  (Exp.  33)  add  NH4OH  to  alkaline 
reaction,  heat  to  boiling  and  add  (NH^oCOs  in  excess. 
Note  the  precipitation  of  SrCOs  and  CaCOs.  Filter, 
wash  the  precipitates  and  dissolve  them  by  pouring  a 
small  quantity  of  hot  HC2H3O2  repeatedly  through  the 
filter.  Write  equations. 

Experiment  35. — Dilute  the  solutions  (Exp.  34)  to 
20—25  cc.,  heat  to  boiling  and  add  ammonium  sulphate 
[(NH4)2SO4]  solution.  Boil  for  two  to  three  minutes  and 
filter.  To  the  filtrates  add  NH4OH  to  alkaline  reaction 
and  then  ammonium  oxalate  [(NH4)2C2O4]  solution.  Note 
the  precipitation  of  CaC2O4.  (Difference,  detection  of 
Ca.)  Write  equations. 

NOTES. — The  solubility  of  the  sulphates  of  strontium  and  calcium 
in  100  cc.  of  water  at  18°  is  n  and  200  mgs.  respectively.  The  correspond- 
ing solubility  of  their  oxalates  is  4.6  and  0.56  mgs.  From  this  it  will 
be  noted  that  a  complete  separation  of  strontium  and  calcium  cannot 
be  obtained  by  means  of  their  sulphates,  but  it  is  also  evident  that  there 
will  remain  in  the  nitrate  enough  CaSO4  to  be  easily  detected  with 
(NH4)2C204. 

The  solubility  of  the  salts  of  barium,  strontium  and  calcium  in  mgs. 
per  100  cc.  of  water  at  18°  is  shown  in  the  following  table: 


C03 

Cr04 

S04 

C204 

Ba  

2-3 

0.38 

0.23 

8.6 

Sr  

i  .  i 

120. 

II  . 

4.6 

Ca 

i  TI 

4OO 

2OO 

o  56 

OUTLINES  OF  ANALYSIS  (TABLE  IX) 


TABLE  IX 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  IV 
Ions  present  in  acid  solution 


;?J 

Reagent 

Ba+  + 

Sr+  + 

Ca+  + 

I 

NH4OH  +  (NH4)2C03 

BaCO3 

SrC03 

CaCO3 

2 
^ 

HC2H3O2 

(a)   CaSO4(sat.  soln.) 

Ba(C2H3O2)2 
BaSO4 

Sr(C2H302)2 
SrSO4 

Ca(C2H2O2)2 
Ca(C2H3O2)2 

(&)   K2Cr207 

BaCrO4 

SrCrO4 

CaCrO4 

4 

NH4OH  +  (NH4)2C03 

1 

SrCO3 

CaCO3 

5 
6 

HC2H3O2 
(a)   CaSO4  (sat.  soln.) 

Sr(C2H302)2 
SrS04 

Ca(C2H3O2)2 
Ca(C2H302)2 

(&)    (NH4)2S04 

SrSO4 

CaSO4 

7 

(NH4)2C204 

1 

CaC2O4 

1 

ANALYSIS 

Group  IV 

Ba++,  Sr++,  Ca+  + 

(80)  Precipitation. — Evaporate  the  filtrate  from  Group 
III   (50)  to  15-20  cc.     Filter  off  any  free  sulphur.     (See 
Discussion  49.)     Heat  to  boiling  (see  Discussion  50),  add 
(NH^COs  as  long  as  a  precipitate  continues  to  form, 
and  allow  the  mixture  to  stand  about  ten  minutes.     A 
white    crystalline    precipitate    indicates    the    presence    of 
Group  IV.     Filter,  treat  the  precipitate  by  (81)  and  the 
filtrate  by  (82).     (See  Discussion  51.) 

(81)  Detection   of  Barium   and   Strontium. — Dissolve 
the  precipitate   (80)   by  pouring  a  hot    lo-cc.   portion  of 
6N.  HC2HsO2  repeatedly  through  the  filter.     To  a  small 
portion  of  the  resulting  solution  add  an  equal  volume  of 
saturated  CaSCU  solution.     The  immediate  formation  of 
a   white   precipitate   shows    the    presence   of   barium.     A 
white  precipitate  which  forms  only  slowly  or  on  warming 
shows  the  absence  of  barium  and  the  presence  of  strontium. 


76  CATIONS— (METAL  IONS) 

If  no  precipitate  forms,  both  barium  and  strontium  are 
absent.     (See  Discussion  51  and  52.) 

If  barium  is  shown  to  be  present,  treat  the  remainder 
of  the  solution  by  (83) ;  if  barium  is  absent  and  strontium 
is  shown  to  be  present,  treat  by  (85) ;  if  both  barium  and 
strontium  are  absent  treat  by  (86). 

(82)  Detection  of  Traces  of  Barium  and  Calcium. — 
To  a  portion  of  the  nitrate  (80)  add  1-2  cc.  of  (NH4)2SO4 
and    warm.     A   white   precipitate    (turbidity)    shows   the 
presence  of  barium.     To  the  remainder  of  the  nitrate  (80) 
add   1-2  cc.  of   (NH4)2C2O4.     A  white  precipitate   (tur- 
bidity) shows  the  presence  of  calcium.     Reunite  the  two 
portions,  filter  off  any  precipitate  and  treat  the  filtrate  for 
Group  V  by  (90).     (See  Discussion  51.) 

(83)  Separation  of  Barium. — Dilute  the  remainder  of 
the    solution    (81)    to    about    75   cc.    and    add   5   cc.   of 
NH4C2H3O2.     Heat  to  boiling,  and  while  hot  add  K2Cr2O7 
in  slight  excess.     A  large  excess  should  be  avoided.     (See 
note,   Exp.   33.)     Filter  while  hot  and   treat  the  filtrate 
by  (84). 

A  confirmatory  test  for  barium  may  be  made  as  follows: 
Dissolve  the  precipitate  of  BaCrO4  in  dilute  HC1;  add 
about  0.5  cc.  of  alcohol  (C2H5OH)  and  boil  until  the  yellow 
color  entirely  disappears  and  the  solution  becomes  green. 
Neutralize  with  NH4OH  and  filter  off  the  Cr(OH)3. 
Acidify  the  filtrate  with  HC1  and  evaporate  nearly  to  dry- 
ness.  Introduce  a  little  of  this  solution  on  a  clean  platinum 
wire  into  the  colorless  flame  of  the  Bunsen  burner.  A 
green  color  shows  the  presence  of  barium. 

(84)  Precipitation  of  Strontium  and  Calcium. — To  the 
filtrate   (83)   add   NH4OH  to  alkaline  reaction;    heat  to 
boiling   and   add    (NH4)2COa   solution   in   excess.     Allow 
the  mixture  to  stand  five  to  ten  minutes  and  filter.     Reject 
the  filtrate.     Dissolve  the  precipitate  in  a  few  cc.  of  6N. 
HC2HsO2  and  test  for  strontium  as  outlined  in  (81). 

(85)  Separation    of    Strontium. — Dilute    the    solution 
(84)  to  about  20  cc.;   heat  to  boiling  and  add  (NH4)2SO4 


DISCUSSION  (49-53)  W 

solution,  as  long  as  a  precipitate  continues  to  form.  Filter 
and  treat  the  nitrate  by  (86). 

A  confirmatory  test  for  strontium  may  be  made  as 
follows:  Add  to  the  precipitate  of  SrSO4  about  ten  times 
its  volume  of  solid  Na2COs  and  10  cc.  of  water,  and  boil 
three  to  five  minutes.  Filter,  dissolve  the  residue  in  2-3 
cc.  of  HC1  and  introduce  a  small  portion  of  the  solution 
on  a  clean  platinum  wire  into  the  colorless  flame  of  the 
Bunsen  burner.  A  deep  red  color  shows  the  presence 
of  strontium.  (Caution:  Do  not  confuse  with  calcium, 
yellowish-red  color.) 

(86)  Detection  of  Calcium.— To  the  filtrate  (85)  add 
NH4OH  to  alkaline  reaction,  and  then  (NH4)  zC&i. 
A  white  crystalline  precipitate  shows  the  presence  of 
calcium. 

A  confirmatory  test  for  calcium  may  be  made  by 
moistening  the  CaC2O4  precipitate  with  a  little  cone. 
HC1  and  introducing  a  small  portion  of  it  on  a  clean 
platinum  wire  into  the  colorless  flame  of  the  Bunsen 
burner.  A  yellowish-red  color  shows  the  presence  of 
calcium. 

DISCUSSION 

49.  The  solution  to  be  tested  for  Group  IV  should 
be  clear  and  colorless  and  should  contain  only  a  moderate 
excess  of  NH4C1.  Any  yellow  or  brown  color  due  to 
undecomposed  (NH4)2S  or  small  quantities  of  nickel  or 
chromium,  not  precipitated  in  Group  III,  will  usually 
be  removed  by  the  evaporation  as  directed  in  (80).  NiS 
is  slightly  soluble  in  excess  (NH4)2S,  forming  a  brown 
solution.  A  large  excess  of  NH4C1  will  hold  chromium 
in  solution  as  CrCls^NHs,  a  red  solution. 

The  filtrate  from  Group  III  may  also  contain  small 
quantities  of  aluminium,  owing  to  its  amphoteric  (see 
Introduction  16)  nature.  Evaporation  decomposes  (NH4)2S 
and  volatilizes  excess  NH3  which  causes  the  precipitation 


78  CATIONS— (METAL  IONS) 

of  any  NiS  and  A1(OH)3  that  may  be  present.  If  the 
solution  is  still  colored  from  chromium  it  should  be  evapo- 
rated to  dryness,  taken  up  in  10  cc.  of  water  and  filtered 
to  remove  excess  ammonium  salts  and  Cr(OH)3. 

50.  Group  IV  should  be  precipitated  from  a  hot  solu- 
tion in  order  to  insure  complete  separation  from  magnes- 
ium, which  forms  a  double  salt,  MgCO3-  (NH4)2CO-4H2O, 
with   (NH4)2CO3  only  moderately  soluble  in  cold  water. 
It  is  much  more  soluble  in  hot  water.     The  solution  should 
not  be  boiled  after  the  addition  of  the  precipitating  agent 
(NH4)2CO3,  owing  to  the  fact  that  it  decomposes  quite 
readily  when  heated.     From  a  cold  solution  the  carbon- 
ates  are   precipitated   in   a   fine   state   of   division.     Heat 
favors  the  formation  of  larger  particles. 

51.  Owing  to  the  appreciable  solubility  of  the  alkali 
earth    carbonates    (see    table    following    note,    Exp.    35), 
traces  of  these   metals   may   remain   in   the   filtrate  after 
precipitation  with  (NH4)2CO3.     The  insolubility  of  their 
phosphates    makes   it   necessary   to   remove   them    before 
testing  for  magnesium  in  Group  V. 

52.  The  method  of  analysis  as  outlined  does  not  permit 
a  complete  separation  of  barium,  strontium  and  calcium; 
yet  it  enables  the  analyst  to  determine  with  a  rather  high 
degree  of  accuracy  their  presence  or  absence.     The  principle 
involved    in    the   detection    of   barium    and    strontium    is 
based  on  the  relative  solubility  of  their  sulphates.      (See 
table    following    notes,    Exp.    35.)     A    saturated    solution 
of  CaSO4  produces  enough  SO4=  to  cause  an  immediate 
precipitation  of  BaSO4,  but  owing  to  the  larger  solubility 
product  of  SrSO4  the  solution  is  saturated  with  it  only 
slowly  or  after  heating.     The  difference  in  time  required 
for  the  precipitation  of  BaSO4  and  SrSO4  under  the  con- 
ditions stated  is  quite  marked,  so  that  it  is  comparatively 
easy  to  distinguish  between  barium  and  strontium. 

53.  The    test    for    calcium     assumes    the    absence    of 
strontium  and  barium  and  the  test  for  strontium  assumes 
the  absence  of  barium;    hence,  if  barium  is  shown  to  be 


PRELIMINARY  EXPERIMENTS  79 

present  it  must  be  removed  before  making  the  tests  for 
strontium  or  calcium.  Precipitation  as  BaCrO4  effect- 
ively removes  the  barium,  while  little  or  no  SrCrO4  or 
CaCrO4  are  precipitated.  (See  table  following  notes, 
Exp.  35-) 

The  separation  of  strontium  and  calcium  with 
(NH4)2SO4  is  not  complete.  Some  of  the  calcium  may 
be  precipitated  with  the  strontium  and  some  of  the  stron- 
tium will  remain  in  the  solution,  though  not  enough  to 
interfere  with  the  test.  Enough  calcium  will  always 
remain  in  the  nitrate,  however,  to  give  a  precipitate 
with  (NH4)2C2O4,  owing  to  the  very  slight  solubility 
of  CaC2O4. 


V 

Preliminary  Experiments 
Mg++,  NH4+,  K+,  Na+ 

In  connection  with  the  following  experiments  study 
Table  X. 

Experiment  36. — To  5  cc.  of  the  test  solution  contain- 
ing Mg++  add  an  equal  volume  of  NH4C1.  Make  dis- 
tinctly alkaline  with  NH4OH  and  add  sodium  phosphate 
(Na2HPO4)  solution.  The  precipitate  is  MgNH4PO4. 
To  a  second  5-cc.  portion  of  the  test  solution  add  NH4OH 
to  alkaline  reaction.  Note  the  precipitation  of  Mg(OH)2. 
Now  add,  without  filtering  an  excess  of  NH4C1.  Why 
does  not  Mg(OH)2  precipitate  in  Group  III?  Explain. 
Write  equations. 

Experiment  37. — Evaporate  a  small  quantity  of  NH4C1 
to  dryness  in  a  porcelain  dish  and  heat  the  residue.  To  a 
second  portion  in  a  test-tube  add  NaOH.  Note  the  odor 
of  the  gas  evolved  and  its  effect  upon  a  piece  of  moist 
red  litmus  paper.  What  is  the  gas  evolved?  Why  can 
this  reaction  be  used  as  a  test  for  NH4+? 


80  CATIONS— (METAL  IONS) 

Experiment  38. — To  separate  i-cc.  portions  of  the  test 
solutions  of  Mg++,  K+  and  Na+  add  10-15  drops  of 
perchloric  acid  (HC1O4).  Allow  the  mixtures  to  stand 
for  a  short  time.  Note  the  precipitation  of  KCIO-*.  (Dif- 
ference, detection  of  K.) 

Experiment  39. — To  separate  i-cc.  portions  of  the  test 
solutions  of  K+  and  Na+  add  an  equal  volume  of  alcohol 
and  then  about  i  cc.  of  fluoboric  acid  (HBF4).  Note 
the  precipitation  of  KBF4.  (Difference,  separation  of  K.) 
Write  equations. 

Experiment  40. — Repeat  Exp.  39,  substituting  fluo- 
silicic  acid  (H2SiF6)  for  the  HBF4.  Note  the  character 
of  the  precipitates  formed.  Why  must  potassium  be 
removed  before  testing  for  sodium?  Write  equations. 

Experiment  41. — Determine  the  color  of  the  flame 
produced  by  potassium  and  sodium  by  introducing  a 
salt  of  each  on  a  clean  platinum  wire,  into  the  colorless 
flame  of  the  Bunsen  burner. 


OUTLINES  OF  ANALYSIS  (TABLE  X) 


81 


TABLE  X 
OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  V 


No. 

Reagent 

Mg+  + 

NH4+ 

K+ 

Na+ 

i 

2 

3 
4 

Na2HPO4 
NH4OH 
NaOH 
(a)  HC1O4 
(b)  HBF4 
H2SiF6 

MgNH4PO4 

NH4  + 
NH,*t 

K+ 
K+ 
KC1O4 

Na+ 
Na+ 
NaClO4 
NaBF4 
Na2SiF« 

1 

KBF4 

i 

ANALYSIS 

Group  V 
Mg++,  NH4+,  K+,  Na+ 

(90)  Detection  of  Magnesium. — Evaporate  the  filtrate 
from  (82)  in  a  porcelain  dish  until  the  ammonium  salts 
begin  to  crystallize  out.     Filter,  and  to  one-third  of  the 
filtrate  add  1-2  cc.  of  Na2HPO4  and  enough  i$N.  NH4OH 
to  make  one-third  the  total  volume.     Shake  the  mixture 
vigorously  for  two  to  three  minutes,  and  if  a  precipitate 
does  not  form,  rub  the  walls  of  the  vessel  with  a  glass  rod 
(see  note,  Exp.  I4a)  and  allow  it  to  stand  for  some  time. 
A  white  crystalline  precipitate  is  MgNH4PO4.     (See  Dis- 
cussion   54.)     Acidify    the    remaining   two-thirds   of   the 
filtrate  above  with  HC1  and  treat  by  (91). 

If  the  precipitate  of  MgNH4PO4  is  of  doubtful  form 
it  should  be  filtered,  dissolved  off  the  filter  with  a  small 
quantity  of  HC2HsO2  and  reprecipitated  by  the  addition 
of  NH4OH  and  a  small  quantity  of  NaHPO4. 

(91)  Detection  of  Potassium. — Transfer   the   solution 
(90)  to  a  porcelain  dish  and  evaporate  to  dryness.    Place 
the  dish  on  a  clay  triangle  and  ignite  till  the  ammonium 
salts  are  completely  expelled.     (See  Discussion  55.)     Dis- 


82  CATIONS— (METAL  IONS) 

solve  the  residue  in  5-6  cc.  of  water,  transfer  about  I  cc. 
of  the  resulting  solution  to  a  wa,tch  glass  and  add  a  few 
drops  of  perchloric  acid  (HC1O4).  The  formation  of  a 
white  crystalline  precipitate  shows  the  presence  of  potas- 
sium. Confirm  by  means  of  the  flame  test.  (See  Pre- 
liminary Exp.  41.)  Treat  the  remainder  of  the  solution 
by  (92)  if  potassium  was  found,  otherwise  by  (93). 

(92)  Removal    of    Potassium. — To    the    remainder    of 
the  solution  (91)  add  an  equal  volume  of  alcohol  and  an 
excess  of  fluoboric  acid  (HBF4).     Filter  off  the  precipitate 
of  KBF4  and  treat  the  filtrate  by  (93). 

(93)  Detection  of  Sodium. — To  the  nitrate   (92)   add 
1-2  cc.  of  fluosilicic  acid  (H2SiF6)  and  allow  the  mixture 
to  stand  a  few  minutes.     (See  Discussion  58.)     A  white 
gelatinous    precipitate    shows    the    presence    of    sodium. 
Confirm  by  means  of   the   flame   test.      (See    Preliminary 
Exp.  41.) 

(94)  Detection   of   Ammonium. — The   test   for   NH4+ 
must  be  made  on  a  small  portion  of  the  original  substance. 
Introduce  3-4  cc.  of  the  original  substance,  if  it  is  a  liquid, 
or  about  o.i  gm.  if  it  is  a  solid,  into  a  small  beaker,  and 
add   NaOH  till  the  mixture  is  distinctly  alkaline.     Test 
the  vapors  with  a  piece  of  red  litmus  paper  placed  on  the 
under  side  of  a  watch  glass  covering  the  beaker.     If  no 
change  is  observed  in  the  litmus,  warm  the  mixture  gently, 
but  do  not  boil.     If  the  red  litmus  turns  blue  ammonium 
salts  are  present. 

DISCUSSION 

54.  Since  there  are  no  characteristic  color  reactions 
that  can  be  used  for  the  detection  of  magnesium  it  becomes 
necessary  to  use  considerable  care  in  manipulation  in 
order  to  secure  a  precipitate  that  can  be  recognized  and 
depended  upon.  MgNH4PO4  is  a  white  crystalline  sub- 
stance soluble  in  HC2H3O2,  but  insoluble  in  NH4OH. 

Practically  all  the  metals,  except  alkali  metals,  form. 


DISCUSSION  (54-58)  83 

phosphates  insoluble  in  NH^OH,  hence  the  necessity  of 
having  a  clear  solution  entirely  free  from  metals  of  the 
previous  groups.  The  presence  of  NH^Cl  is  necessary 
to  prevent  the  precipitation  of  Mg(OH)2,  a  white  flocculent 
precipitate.  (See  Discussion  34.) 

55.  Since  the  presence  of  ammonium   salts  interferes 
with  the  subsequent  tests  for  potassium  and  sodium,   it 
is  necessary  that  their  removal  be  complete.     Ammonium 
compounds  react  with  both  HC1O4  and  H^SiFe  to  form 
the  corresponding  ammonium  salts,  which  are  only  slightly 
soluble  under  the  conditions  of  the  experiments.     In  order 
to   insure  the  complete  volatilization   of  the  ammonium 
salts  all  parts  of  the  dish  should  be  well  heated,  though 
it  must  not  be  heated  nearly  to  redness  since  both  KC1 
and  NaCl  are  somewhat  volatile  at  that  temperature. 

56.  In  testing    for    potassium    and  sodium  by  means 
of  the  flame,  a  faint  yellow  coloration  will  almost  invari- 
ably be  obtained,  owing  to  the  fact  that  traces  of  sodium 
are  nearly  always  present  in  the  reagents  previously  used. 
A  fleeting  yellow  tinge  should  not  be  taken  as  evidence 
of  the  presence  of  sodium,  but  the  yellow  color  should  be 
distinct  and  persistent. 

57.  The   color  of   the   flame   may   be   used   to   detect 
potassium    in    the    presence  of  considerable  amounts  of 
sodium,  if  a  blue  glass  is  used  to  cut  out  the  yellow  sodium 
rays.     With  small  amounts  of  potassium,   however,   this 
test  is  not  always  satisfactory. 

58.  Fluosilicic  acid  (H2SiF6)  reacts  slowly  with  glass, 
so  that  on  long  standing  in  a  glass  vessel  it  will  usually 
produce   a   precipitate   even    in    the   absence    of    sodium. 
Since    the    Na2SiF6    precipitate     is    semitransparent,    the 
reaction  should  be  carried  out  in  a  clear  test-tube  or  on 
a  watch  glass. 


84  CATIONS— (METAL  IONS) 

QUESTIONS  FOR  REVIEW 

1.  Name  the  group  reagents  and  the  compounds  pre- 
cipitated by  each. 

2.  Why   is  a  precipitate   sometimes   obtained   on   the 
addition  of  water  before  adding  the  group  reagent?    Give 
examples. 

3.  Why    do    bismuth    and    antimony    sometimes    pre- 
cipitate in  Group  I  ? 

4.  What   is   the   effect   of   adding   NH4OH    to   AgCl, 
Hg2Cl2?     Write  equations. 

5.  Can  HC1  be  substituted  for  NH4C1— HNO3  in  the 
precipitation  of  Group  I?     Explain.     Could  NaCl  be  used? 

6.  Explain    what    is    meant   by   the    following    terms: 
hydrolysis,  reagent,  precipitate  residue,   filtrate,   colloidal 
solution. 

7.  Define  ion,  acid,  base,  salt,  oxidation,  reduction. 

8.  State  the  law  of  mass  action. 

9.  What  is  meant  by  common  ion,  solubility  product, 
ionization  constant,  complex  ion? 

10.  How  may  2N.  HC1  be  made  from  6N.  HC1? 

11.  Explain  by  means  of  the  solubility-product  principle 
why  PbCb  is  less  soluble  in  a  solution  containing  NH4C1 
than  in  pure  water. 

12.  Explain  by  means  of  the  law  of  mass  action  why 
AgCl  is   soluble   in  NH4OH  and  is  reprecipitated  on  the 
addition  of  an  acid. 

13.  Why  is  it  necessary  to  have  a  definite  concentration 
of  acid  for  the  precipitation  of  Group  II?     What  is  the 
effect  if  the  acid  concentration  is  too  great,  or  if  it  is  too 
small?     Explain  by  means  of  the  solubility-product  prin- 
ciple. 

14.  What  precautions  must  be  taken  in  the  treatment 
of  the  Group  II  precipitate  with  (NH4)2S2  reagent!     Why? 

15.  Why  is  it  necessary  to  test  for  lead  and  mercury 
in  both  Groups  I  and  II? 

1 6.  What  are  the  colors  of  the  following  compounds, 


QUESTIONS  FOR  REVIEW  85 

precipitated   in  Group   II:   HgS,    Pbs,    Bi2S3,   CuS,  CdS, 
As2S3,  Sb2S3,  SnS,  SnS2? 

17.  Why  is  it  necessary  to  evaporate  till  white  fumes 
appear  in  the  separation  of  lead  in  Group  II? 

1 8.  What  is  the  effect  of  adding  NH4OH  to  a  solution 
of  CuSO4  and  Bi2(SO4)2?     Write  equations.     Explain. 

19.  What  is  the  purpose  of  KCN  in  the  detection  of 
cadmium?     Write  equations  involved. 

20.  Explain  by  means  of  the  solubility-product  prin- 
ciple why  CuS,  which  is  only  slightly  soluble  in  hot  2N. 
HC1,  dissolves  readily  in  hot  2N.  HNO3. 

21.  What  effect  does  temperature  have  on  the  complete- 
ness with  which  Group  II  sulphides  may  be  precipitated? 
Explain. 

22.  What  precautions  are  necessary  to  insure  complete 
precipitation  of  arsenic  in  Group  II?     Explain. 

23.  Given   a  solution  which  is  known   to   contain   no 
other  metals  than  those  given  below,  outline  a  method 
of  analysis  that  will  necessitate  no  unnecessary  steps: 

(a)  Lead,  mercury,  cadmium. 

(b)  Copper,  arsenic,  cadmium. 

(c)  Bismuth,  cadmium,  antimony,  tin. 

24.  What  is  the  purpose  of  H2C2O4  in  the  detection 
of  antimony? 

25.  Show  by  a  series  of  equations  the  changes  through 
which  arsenic  goes  when  H2S  is  passed  through  a  dilute 
HC1  solution  of  H3AsO4. 

26.  Explain  by  means  of  the  solubility-product  prin- 
ciple why  PbSC>4  may  be  dissolved  in  NH4C2H3O2  solution. 

27.  Should  PbCrO4  dissolve  in  NH4C2H3O2  solution? 
Explain.     Why  does  PbCrCU  precipitate  from  the  same 
solution  that  dissolves  PbSC>4? 

28.  How    is    Na2SnO2     prepared?      Write    equations 
showing  all  the  changes  that  take  place. 

29.  Explain  by  means  of  the  law  of  mass  action  why 
the  addition  of  HC1  to    a    solution  of   (Na4)2SnS3  pre- 
cipitates SnS2. 


86  CATIONS—  (METAL  IONS 


30.  In   the   separation  of  As2S5  from  80285  and  SnS2 
with  I2N.  HC1  why  does  more  As2S5  dissolve  if  the  solu- 
tion is  allowed  to  boil? 

31.  Why  use  just  10  cc.  of  I2N.  HC1  in  the  treatment 
of  the  sulphides  of  the  tin  division? 

32.  What   is   the    principle   on   which    the   separation 
of    antimony    and    tin,    by    precipitation    with    H2S,    is 
based?      Explain    by    means     of    the     solubility-product 
principle. 

33.  What  is  the  confirmatory  test  for  tin?     What  is 
the  precipitate  formed?     Write  all  equations  involved. 

34.  If   phosphate   is  present,   what  elements   may   be 
precipitated  on  the  addition  of  the  third  group  reagent? 

35.  Why  is  it  possible  to  separate  the  Al  division  from 
the  Fe  division  with  NaOH  and  Na2O2? 

36.  What  elements  are  affected  by  the  Na2C>2?     Write 
equations. 

37.  Why  is  Na2COs  added  in  the  separation  of  the 
Al  and  Fe  divisions? 

38.  Why  is  HNO3  added  before  NH4OH  in  the  sepa- 
ration of  aluminium? 

39.  What  is  the  purpose  of  H2O2  in  dissolving  the  Fe 
division  ? 

40.  If  it  is  necessary  to  use  HNOs  to  dissolve  Group 
III  what  conclusion  can  be  drawn? 

41.  Since  nickel  and  cobalt  sulphides  are  not  precipi- 
tated in  acid  solution  why  is  it  necessary  to  use  aqua  regia 
to  dissolve  them? 

42.  Given   a  solution  which   is  known   to   contain   no 
other  metals  than  those  given  below,  outline  a  method 
of  anlysis  that  will  necessitate  no  unnecessary  steps; 

(a)  Aluminium,  chromium,  iron. 

(b)  Chromium,  manganese,  cobalt,  nickel. 

(c)  Zinc,  manganese,  iron,  nickel. 

43.  Why  is  it  necessary  to   test  for  zinc  in   the   Fe 
division? 

44.  If  phosphate  is  present  in  an  unknown,  in  what 


QUESTIONS  FOR  REVIEW  87 

solutions  is  it  necessary  to  test  for  the  metals  of  Group  IV? 
Why? 

45.  What  is  the  purpose  of  the  following  reagents  in 
the  analysis  of  Group  IV;    CaSO4,  K2Cr2O7,    (NH4)2SO4, 
(NH4)2C204+? 

46.  Could  K2CrO4  be  used  instead  of  K2Cr2O?  in  the 
precipitation  of  barium  and  lead?     What  is  the  relation 
between  K2CrO4  and  K2Cr2O?? 

47.  Why    is    it    necessary    to    add     (NH4)2SO4    and 
(NH4)2C2O4  to  the  filtrate  from  Group  IV  before  testing 
for  Group  V? 

48.  Why  does  Mg(OH)2  not  precipitate  in  Group  III 
or  Group  IV?     Explain  by  means  of  the  solubility-product 
principle. 

49.  Why  not  evaporate  to  dry  ness  and  drive  off  all 
ammonium  salts  before  testing  for  magnesium? 

50.  Why  is  it  necessary  to  test  the  original  material 
for  NH4  +  ? 

51.  Why  remove  ammonium  salts  before  testing  for 
potassium  and  sodium? 

52.  What  is  the  purpose  of  HBF4  in  the  analysis  of 
Group  V? 

53.  What  salts  are  most  suitable  for  use  in  flame  tests? 
Why? 

54.  If  an  unknown  is  soluble  in  water,  and  phosphate 
is  found,  what  metals  will  it  be  unnecessary  to  test  for? 


PART  III 


ACIDS 

In  most  cases  the  acid  ions  (anions)  exist  in  solution 
as  compound  radicals  composed  of  two  or  more  elements 
held  together  and  acting  as  a  single  substance.  In  this 
respect  they  differ  from  the  metal  ions  (cations)  which 
usually  exist  in  solution  as  simple  radicals.  The  most 
important  anions  consisting  of  a  single  element  are  the 
halides  (Cl~,  Br~,  I~,  F~)  and  sulphide  (S=).  Because 
of  this  difference,  more  care  must  be  exercised  not  only 
in  the  preparation  of  the  solution  for  the  detection  of  acids 
but  also  in  the  treatment  during  analysis  in  order  to  pre- 
vent decomposition,  oxidation  or  reduction. 

The  anions  for  whose  separation  and  detection  pro- 
vision is  made  in  this  scheme  of  analysis  are : 

Group  I. — Anions  whose  silver  salts  are  insoluble  in 
cold  dilute  HNO3. 

Ferrocyanide,    Fe(CN)e~    ~~      Sulphide,  S~ 

Ferricyanide,     Fe(CN)6~  Iodide,  I~ 

Thiocyanate,     CNS~  Bromide,  Br~ 

Cyanide,  CN~  Chloride,  Cl~ 

Group  II. — Anions  whose  salts  decompose  on  boiling 
in  acid  solution  and  give  characteristic  volatile  oxides. 
Carbonate,         CO3~  Thiosulphate,  S2O3~ 

Sulphite,  SO3~  Nitrite,  NO2~ 

Group  HI. — Anions  whose  silver  salts  are  soluble  in 
acid  but  insoluble  in  warm  neutral  solution. 

Arsenite,  AsO3 Arsenate,          AsO4 

Oxalate,  C2O4~  Phosphate,        PO4- 

Chromate,          CrO4~  Tartrate,  C4H4Oe — 

88 


THE  SYSTEMATIC  ANALYSIS  89 

Group  IV. — Anions  whose  silver  salts  are  soluble. 
Sulphate,  SO4~  Acetate,  C2H3O2- 

Borate,  BO3-  Nitrate,  NO3- 

Fluoride,  F~ 

Provision  was  made  for  the  detection  of  silicate  (SiO3~~) 
in  the  course  of  the  analysis  for  metal  ions.     (See  (5)  ). 


THE  SYSTEMATIC  ANALYSIS 
ANIONS 

Since  any  interfering  cations  will  be  removed  during 
the  course  of  the  analysis,  the  solution  as  prepared  for 
the  metal  analysis  may  be  used  for  the  analysis  of  anions 
if  the  substance  has  been  dissolved  in  water  or  cold  dilute 
HNO3.  If  the  substance  is  a  liquid  or  solution  treat 
directly  by  (no);  if  a  solid  treat  by  (100). 

PREPARATION  OF  SOLUTION 

(100)  Treatment  of  a  Solid. — To  about  0.5  gram  of 
the  finely  powdered  substance  add   10  cc.  of  cold  water 
and   mix  thoroughly.     Filter  and  wash   the  residue  with 
5  cc.  of  cold  water,  catching  the  wash  water  in  the  vessel 
containing  the  nitrate.     Acidify  the  filtrate  with  HNO3, 
noting  if  a  gas  is  evolved  (see  Discussion  60),  and  treat 
by  (no).     Treat  the  residue  by  (101). 

(101)  Treatment  with  Dilute  HNO3. — Pour  repeatedly 
through  the  filter  containing  the  residue  undissolved  by 
cold   water,   a  5-cc.   portion  of  cold   2N.   HNO3,    noting 
if  a  gas  is  evolved.     (See   Discussion   59.)     If  a  residue 
remains,  punch  a  hole  through  the  filter  and  wash  it  into 
a  test-tube  with  5  cc.  of  6N.  HNO3.     Warm  the  mixture 
as  long  as  any  of  the  solid   seems  to.  dissolve.     Dilute 
with  5  cc.  of  water,  cool  and  filter.     Treat  the  combined 
filtrates  by  (no).     Treat  the  residue  by  (102). 


90  ACIDS 

(102)  Treatment  with  Na2CO.3. — If  there  is  a  residue 
undissolved  by  dilute  HNO3,  the  anion  constituents  may 
be  dissolved  by  treatment  with  Na2CO3  solution,  as 
directed  in  (6)  or  by  fusion  with  Na2CO3  as  directed  in 
(7).  In  either  case,  however,  6N.  HNO3  should  be  sub- 
stituted for  HC1,  and  the  solution  boiled  to  drive  off  the 
CO2  before  treatment  with  AgNO3  by  (no). 

DISCUSSION 

59.  A  careful  study  of  the  solubility  (see  table,  Appen- 
dix,  III)  and  acidity  of  the  material  to  be  analyzed,  in 
connection  with  the  metal  ions  found,  will  often  permit 
the  omission  of  certain  steps  in  the  system  of  analysis 
for  anions.     Thus,  if  the  substance  contains  silver  and  is 
soluble  in  water  or  dilute  acid,  it  is  unnecessary  to  test  for 
Group    I.     Barium    or    lead    and    sulphate    cannot    exist 
together  in  a  neutral  water-soluble  material.     Group    II 
anions  will  not  be  present  in  a  strongly  acid  solution. 

A  study  of  the  nature  and  color  of  the  group  precipi- 
tates may  also  permit  the  omission  of  certain  steps;  e.g., 
Ag2S  is  black;  Ag3Fe(CN)6,  Ag2CrO4  and  Ag3AsO4  are 
dark  red;  Agl,  Ag3PO4  and  Ag3AsO3  are  yellow;  AgBr 
is  yellowish-white,  while  the  remaining  silver  salts  of 
Groups  I  and  III  are  white. 

60.  When  a  strong  mineral  acid  is  added  to  a  carbonate, 
the  carbonic  acid   formed    (H2CO3)   immediately  decom- 
poses and,  owing  to  its  slight  solubility,  CO2,  is  evolved. 
Nitrites,  sulphites  and  thiosulphates  also  tend  to  decom- 
pose in  acid  solution  with  the  formation  of  N2O3,  SO2, 
and   SO2+S   respectively.     The   solubility   of    N2O3    and 
SO2  is  so  great,  however,  that  enough  will  always  remain 
in  solution  for  their  detection  according  to    the  methods 
outlined  in  Group  II. 

If  the  evolution  of  a  gas  is  noted  during  the  treatments 
as  directed  in  (100)  or  (101),  and  carbonate  is  not  found 
in  the  solution  as  prepared,  a  small  portion  of  the  orginal 
unknown  should  be  tested  for  carbonate  as  directed  in  (130). 


OUTLINES  OF  ANALYSIS  (TABLE  XI) 


91 


TABLE  XI 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  I 

(DIVISION  A) 

Ions  present  in  dil.  HNO3  solution 


No. 

Reagent 

Fe(CN)6= 

Fe(CN)6= 

CNS-               CN- 

i 

i 

2 

3 

4 

5 
6 

7 

AgN03 
NaCl—  HC1 

(NH4)2MoO4 

Zn(N03)2 
HC1 
FeSO4 
FeCl3 
NaOH— 
FeS04 
HC1 

Ag4Fe(CN)6 

Ag3Fe(CN)6 

Ag( 
Na( 
Na( 

Zn( 

J 
Fe( 

, 

:NS 

:NS 
:NS 

CNS)2 

, 

CNS)3 

AgCN 

NaCN 
NaCN 

Zn(CN)2 

Fe(CN)2 

Na4Fe(CN)6 
Fe4(Fe(CN)6)3 

Na4F 
Na»(l 

(Fe(( 
2Mo 

e(CN)4 
tfoO2)3- 

:N),),. 

2.3 

Na3Fe(CN)6 

Na3Fe(CN)6 

Zn3(Fe(CN)6)2 

H3Fe(CN)6 
Fe,(Fe(CN)6)2 

! 

ANALYSIS 

Group  I 
Fe(CN)6=  Fe(CN)6=,  CNS-,  CN-,  S=  I~,  Br~,  Cl- 

(no)  Precipitation. — To  about  25  cc.  of  the  cold  solu- 
tion acidified  with  HNOs  (see  Discussion  60)  add,  slowly 
and  with  constant  stirring,  AgNO3  solution  in  excess. 
Filter,  divide  the  precipitate  into  two  portions,  treat  one 
by  (in)  and  the  other  by  (120).  Reserve  the  filtrate  for 
analysis  of  Group  II  (130). 


Division  A 


(in)  Solution  of  the  Cyanogen  Compounds. — Suspend 
one  portion  of  the  precipitate  (no)  in  5  cc.  of  water,  add 
5  cc.  of  the  NaCl  reagent  and  mix  thoroughly.  Filter, 


92  ACIDS 

reject  the    residue    of    silver  halides   (see  Discussion  61) 
and  treat  the  nitrate  by  (112).     '(See  Discussion  59.) 

(112)  Detection  of  Ferrocyanide. — To  the  nitrate  (in) 
add    ammonium    molybdate    (  (NH4)2MoO4)    reagent    in 
excess.     A  red  flocculent  precipitate  shows  the  presence 
of  ferrocyanide.     (See   Discussion   62.)     Filter  and   treat 
the  nitrate  by  (113). 

(113)  Precipitation   of   Ferricyanide. — To   the    nitrate 
(112)    add   a   slight  excess  of   Zn(NO)a   solution.     Allow 
the  mixture  to  stand  a  few  minutes.     A  white  precipitate 
indicates  ferricyanide.     If  the  precipitate  is  colloidal  or 
finely  divided  shake  the  mixture  with  a  pinch  of  asbestos 
fiber  to  coagulate  the  precipitate  and   filter.     Treat  the 
precipitate  by  (114)  and  the  filtrate  by  (115). 

(114)  Confirmatory   Test   for   Ferricyanide. — Dissolve 
the  precipitate   (113)    by   pouring    a    3-5   cc.   portion    of 
6N.  HC1  repeatedly  through  the  filter.     Dilute  the  solu- 
tion with  an  equal  volume  of  water  and  add  about  i  cc. 
of   FeSCU   solution.     A  deep   blue   precipitate  shows  the 
presence  of  ferricyanide. 

(115)  Detection    of    Thiocyanate    and    Cyanide. — To 
the  filtrate  (113)  add  about  i   cc.  of  FeCl3  solution.     A 
red  coloration  shows  the  presence  of  thiocyanate. 

Make  the  solution  just  alkaline  with  NaOH  and  add 
a  few  drops  of  FeSCU  solution.  Heat  the  mixture  to  boil- 
ing and  boil  gently  for  one  to  two  minutes.  Acidify  with 
HC1  to  dissolve  the  hydroxides  of  iron.  A  blue  residue 
insoluble  in  HC1  shows  the  presence  of  cyanide.  (See 
Discussion  63.) 

DISCUSSION 

61.  The  solution  of  the  cyanogen  compounds  by  means 
of  NaCl  is  based  on  the  relative  solubility  of  their  silver 
salts.  The  lower  solubility  product  of  AgCl  causes  a 
displacement  of  the  equilibrium  toward  the  formation 
of  more  AgCl  and  a  consequent  decrease  of  Ag+,  with  the 


DISCUSSION  (61-63)  93 

result  that  the  ion  product  of  the  cyanogen  compounds 
of  silver  is  reduced  below  their  solubility-product  value 
and  they  pass  into  solution  as  sodium  salts.  In  the  pres- 
ence of  a  small  amount  of  HC1  the  reaction  is  rapid  and 
complete.  The  sulphide  and  other  halides  are  unaffected, 
owing  to  the  fact  that  their  solubilities  are  less  than  that 
of  AgCl.  (See  Introduction  12.) 

62.  The  precipitation  of  ferrocyanide  with  (NHU^MoC^ 
reagent  is  complete  only  when  a  considerable  excess  of 
(NH4)2MoO4  is  present.     Unless  the  ferrocyanide  is  com- 
pletely removed  it  will  be  precipitated  with    ferricyanide 
and  interfere  with  the  test  for  that  acid.     The  heavy  red 
precipitate  of  ferrocyanide  is  a  complex  salt  of  molybdenum 
with  K4Fe(CN)G,  of  the  probable  formula 

K2(MoO2  •  Fe(CN)6)3  •  2MoO3  •  2oH2O, 

and  is  somewhat  soluble  in  excess  of  K4Fe(CN)e.  (Note. — 
The  student  will  learn  to  distinguish  between  ferrocyanide 
and  ferricyanide  by  noting  that  a  ferrocyanide  contains 
a  complex  ion  consisting  of  CN~  and  ferrous  iron  while 
a  ferricyanide  contains  a  complex  ion  consisting  of  CN~ 
and  ferric  iron.) 

63.  When  a  simple  cyanide  is  boiled  with  NaOH  and 
FeSCU,  Na4Fe(CN)e  is  formed.     This  reacts  in  acid  solu- 
tion with  FeCls  to  form  Fe4(Fe(CN)6)3  (Prussian  blue). 
Very  small  quantities  of  this  are  not  readily  detected  in 
the  presence  of  much  thiocyanate,   but  if  the  solution  is 
filtered  a  blue  precipitate  may  easily  be  detected  on  the 
filter  paper. 


94 


ACIDS 


TABLE  XII 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  I 

(DIVISION  B) 

Ions  present  in  dil.  HNOs  solution 


No. 

Reagent 

s- 

I- 

Br~                  Cl  - 

l 

i 

AgN03 

Ag2S 

Agl 

AgBr 

AgCl 

2 

HNO3 

Ag2SO4 

Agl 

AgBr 

AgCl 

HC1 

H2S04 

1 

i 

1 

BaCl2 

BaSO4 

3 

Zn—  HLSO4 

1 

HI 

HBr 

HC1 

4 

K*Cr,O7  (cold) 

I2 

HBr 

HC1 

5 

K2Cr2O7  (hot) 

T 

Br2 

HC1 

KI 

KBrOO 

1 

6 

KMn04 

I    ' 

Cl, 

KI 

KC1— 

ANALYSIS 


Division  B 

(120)  Detection  of  Sulphide. — To  the  second  portion 
(no)  add  5  cc.  of  6N.  HNO3  and  heat  to  boiling  to  oxidize 
the    sulphide    to    sulphate.     Cool    and    filter.     Treat    the 
residue  by  (121).     Evaporate  the  filtrate  nearly  to  dry- 
ness,  add  1-2  cc.  of  6N.  HC1,  filter  and  add  BaCl2  to  the 
clear  filtrate.     A  white  precipitate  of  BaSO4  shows  the 
presence  of  sulphide.     (See  Discussion  64.) 

(121)  Solution  of  the  Halides. — If  any  of  the  cyanogen 
compounds   have   been   found   they  should   be   destroyed 
(see  Discussion  65)  by  placing  the  residue  (120)  in  a  por- 
celain dish  and  igniting  to  dull  redness.     After  the  dish 
and  contents  have  cooled,  add  a  small  piece  of  granulated 
zinc  and  5  cc.  of  6N.  H2SO4.     By  means  of  a  glass  rod 
loosen    any   particles  adhering  to   the    dish   in   order   to 
give  them  free  access  to  the  reducing  agent.     When  the 
reaction   is   complete   and    only   a   black   spongy   residue 


OUTLINES  OF  ANALYSIS  (TABLE  XII) 


95 


remains,   filter,   reject  the  residue  of  metallic  silver  and 
treat  the  filtrate  by  (122). 

(122)  Detection  of  Iodide. — To  the  cold  filtrate  (121) 
add  5  cc.  of  6N.  H2SO4  and   10  cc.  of  water.     Transfer 
the  solution  to  a  test-tube  and  add  about  i  cc.  of  carbon 
tetrachloride  (CCU)  and  a  few  drops  of  K^C^O?  solution. 
Shake    the    mixture    thoroughly.     The  presence  of  iodide 
is  shown  by  a  violet  color  in   the   CCU   layer.     If  iodide 
is  found   filter  and   repeat  the  treatment  with  CCU   and 
K^C^OT  until  the  CCU  layer  shows  no  further  coloration. 
Treat  the  filtrate  by  (123). 

(123)  Detection    of    Bromide. — Transfer    the    filtrate 
(122)  to  a  small  flask  fitted  with  a  delivery  tube  as  shown 
in    Fig.    4.      Place    I    cc.    of   KI 

solution  and  about  10  cc.  of  water 
in  the  receiver  and  connect  to  the 
delivery  tube  as  shown.  Heat 
the  contents  of  the  reaction  flask 
to  boiling  and  boil  for  about  one 
minute.  Remove  the  receiver, 
add  i  cc.  of  CCU  and  shake 
the  contents  vigorously.  A  pur- 
ple coloration  in  the  CCU  layer 
shows  the  presence  of  bromide. 
If  bromide  is  found  the  boiling 
should  be  continued  until  no 
further  test  is  obtained  with  KI 
solution  and  CCU.  Care  should 
be  taken  to  keep  the  concentra- 
tion of  the  acid  in  the  reaction 
flask  near  its  original  value  by 


FIG.  4. 


replacing  with  water  the  loss  due  to  evaporation.  (See 
Discussion  66.)  If  the  yellow  color  should  disappear 
from  the  solution  in  the  reaction  flask  a  few  more  drops 
of  K2Cr2O7  solution  should  be  added. 

(124)  Detection  of  Chloride. — To  the  contents  of  the 
reaction  flask  from  which  all  bromide  has  been  removed 


96  ACIDS 

(123),  add  a  few  drops  of  KMnCX  solution  and  repeat  the 
boiling,  at  the  same  time  catching  the  evolved  gas  in  KI 
solution  as  outlined  in  (123)  and  testing  it  with  CCU  as 
before.  A  purple  color  in  the  CCU  layer  shows  the  pres- 
ence of  chloride. 

DISCUSSION 

64.  By  the  action  of  boiling  HNOs,  according  to  the 
equation 

3H2S  +8HN03-+  3H2S04  +8NO  +4H2O, 

sulphides  are  oxidized  to  sulphates,  which  may  be  detected 
by  the  white  precipitate  of  BaSCU  produced  by  BaCl2. 
Excess  HNOs  should  be  removed  from  the  solution,  how- 
ever, in  order  to  prevent  the  possible  precipitation  of 
Ba(NOs)2,  a  white  crystalline  precipitate,  on  the  sub- 
sequent addition  of  BaCl2.  HC1  is  added  to  remove  any 
silver  ion  that  may  be  in  the  solution. 

The  presence  of  sulphide  may  be  due  not  only  to  its 
presence  in  the  original  material  but  also  to  the  decompo- 
sition of  thiosulphate.  When  AgNOa  is  added  to  an 
acid  solution  containing  a  thiosulphate,  decomposition 
of  the  thiosulphate  takes  place  with  the  formation 
of  insoluble  Ag2S.  When  much  thiosulphate  is  present 
the  decomposition  is  readily  detected  by  more  or  less  rapid 
change  of  color  in  the  precipitate  from  white  through 
varying  shades  of  gray  to  black. 

65.  The    reduction   of   the   silver   halides   to    metallic 
silver  and  the  halogen  acid  by  means  of  zinc  and  H2SO4 
is  complete  only  in  the  absence  of  the  cyanogen  compounds. 
If  any  of  these  compounds  are  found,  therefore,  they  must 
be  removed  before  the  treatment  with  zinc  and  H2SO4. 
They  are  readily  destroyed  by  a  dull  red  heat.     A  higher 
temperature  might  volatilize  some  silver  halide. 

66.  The   separation   and   detection   of   the   halides   is 
based  on  their  relative  resistance  to  oxidation  by  K2Cr2O7 


DISCUSSION  (64-66)  97 

in  acid  solution.  In  a  cold  I-5N.  H2SO4  solution  K2Cr2O7 
readily  oxidizes  HI.  HBr  is  unattacked  by  the  dichromate 
in  acid  concentrations  as  high  as  3N.  if  the  solution  is  kept 
cold,  but  is  oxidized  when  the  solution  is  boiled.  Under 
these  conditions  HC1  remains  unchanged  but  is  readily 
oxidized  by  the  more  powerful  oxidizing  action  of  KMnO4. 
It  is  also  slowly  oxidized  by  R^C^Oy  if  the  acid  con- 
centration is  allowed  to  become  much  greater  than  3  N. 
Hence  the  necessity  of  adding  water,  to  keep  the  solution 
near  its  original  volume  when  testing  for  bromide,  is 
evident. 


98 


ACIDS 


TABLE  XIII 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  II 
Ions  present  in  dil.  HNO3  solution 


No. 

Reagent 

C03= 

S03= 

S203= 

NO2- 

i 

2 
^ 

Ca(OH)2 
HNO3 
HgNO3 

CaCO4 

| 

CaSO3 
H2SO3 
H,SO4(Hg) 

CaSO3 
H2S03 
H2SO4(Hg) 

Ca(N02)2 
HNO2 
HN02 

4 

FeSO4 

i 

I 

I 

FeSO4-NO 

C03=, 

ANALYSIS 
Group  II 
SO3=,  S2O3=,  NO2- 

(130)  Separation  of  Group  and  Detection  of  Carbon- 
ate.— Transfer  the  nitrate  (no)  to  a  small  flask  connected 
with   a  delivery  tube  and   receiver  as  shown  in   Fig.   4. 
About  10  cc.  of  lime-water  (Ca(OH)2)  should  be  placed  in 
the  receiver  so  that  the  delivery  tube  reaches  below  the 
surface  of  the  liquid.     Heat  the  solution  in  the  flask  to 
boiling  and  boil  for  about  one  minute.     If  a  milky  white 
precipitate  forms  in  the  receiver  (see  Discussion  67)  car- 
bonate is  shown  to  be  present.     Filter  and  treat  the  fil- 
trate by   (131).     Reserve  the  solution  in    the    flask    for 
analysis  of  Group  III  (140). 

(131)  Detection  of  Sulphite. — Acidify  the  filtrate  (130) 
with   HNO3  and  add  a  few  drops  of  mercurous  nitrate 
(HgNO3).     A  gray  precipitate  shows  the  presence  of  sul- 
phite.    (See   Discussion  68.)     Filter  and  treat  the  filtrate 
by  (132). 

(132)  Detection  of  Nitrite — To  the  filtrate  (131)    add 
about  i  cc.  of  FeSO4  solution.     The  appearance  of  a  brown 
coloration   shows   the   presence   of   nitrite.     (See    Discus- 
sion 69.) 


DISCUSSION  (69-69)  99 

DISCUSSION 

67.  When  CO2  is  passed  through  lime-water,  insoluble 
CaCOs  is  first  formed.     Excess  of  CO2,   however,  reacts 
with  the  normal   carbonate  to  form   soluble   Ca(HCO3). 
In  the  detection  of  carbonate  the  formation  of  a  white 
precipitate  of  CaCOs,  which  may  or  may  not  redissolve, 
is  evidence  of  the  presence  of  carbonate. 

68.  Mercurous  nitrate  (HgNOs)  is  very  readily  reduced 
to  Hg  by  the  action  of  free  H2SO3,  but  is  unaffected  by 
HNO2.     The  reaction  proceeds  as  follows: 

2HgNO3+H2SO3+H2O^2Hg+H2SO4+2HNO3 

It  has  been  mentioned  (see  Discussion  64)  that  Ag2S2Os 
decomposes  even  in  acid  solution  with  the  formation  of 
Ag2S.  The  reaction  is  as  follows: 


Ag2S2O3  +H2O-+  Ag2S  +H2SO4 

The  decomposition,  however,  is  comparatively  slow,  and 
some  undecomposed  H2S2C>3  may  be  left  in  the  filtrate 
from  Group  I.  Boiling  the  solution  to  expel  Group  II 
decomposes  this  with  the  formation  of  SO2  and  free  sul- 
phur. The  detection  of  both  sulphide  and  sulphite,  there- 
fore, may  have  been  due  to  the  presence  of  thiosulphate 
in  the  original  material.  In  this  case  a  portion  of  the  origi- 
nal solution  should  be  tested  for  sulphide,  thiosulphate 
and  sulphite  as  follows: 

To  the  solution  made  slightly  acid  with  HC2H3O2  add 
ZnSCU  solution.  This  will  precipitate  sulphide  as  white 
flocculent  ZnS.  Filter  and  treat  the  filtrate  with  Sr(NOa)a 
solution  and  allow  the  mixture  to  stand  for  several  hours. 
The  formation  of  a  white  precipitate  (SrSOs)  indicates 
the  presence  of  sulphite.  This  may  be  confirmed  by  the 
property  of  its  HC1  solution  to  decolorize  iodine  solution. 
The  thiosulphate  if  present,  remains  in  the  filtrate  from 
the  SrSOs.  It  can  be  detected  by  acidifying  with  HC1 
and  warming,  when  sulphur  will  be  deposited. 


100  ACIDS 

69.  The   test  for  nitrite  depends  on  the  formation  of 
brown  FeSO4-NO,  according  to  the  following  equations: 


3HN03  +3FeS04  +3HNO2-»  Fe2(SO4)3  +Fe(NO3)3 

+3NO+3H20 
FeS04+NO->FeS04-NO 

If  much  HgNOs  was  added  in  testing  for  sulphite,  a  white 
precipitate  of  Hg2SO4  will  be  formed  on  the  addition  of 
FeSO4. 


102 


ACIDS 


<  o 
2    K 


q 
E 
u 

qq           do                 q 

4^        5  9      }    ^              J    ^       ^ 

II        1      1            18 

k 

u 

of           o         d 

{So  cd                    c3                 cj 

<2               £            £                       « 

III 

2 

o|_o'_|oo    ||_ 

"i 

d  d           d         <  d  ^ 
<  ^;            Jz;         P 

g 

^  ^      _,  d    6        _^ 
^^  ~        ^    4^ 

! 

o|o  ^           _^ 

b/    Sibi  c/3   cB* 

d 

<u 
bo 

1 

<N 

d         d    ^      g 

Sill 

* 

d 

M         „                „                                  * 

OUTLINES  OF  ANALYSIS  (TABLE  XIV)  103 

ANALYSIS 
Group  III 
=  C2O4=,  AsO4=,  PO4=  CrO4=,  C4H4O6- 


(140)  Precipitation.  —  To  the  lukewarm  solution  left  in 
the  flask  (130)  add  NaOH  drop  by  drop  from  a  pipette, 
until,  on  shaking,  the  dark  Ag2O  precipitate  just  fails  to 
redissolve.     (See    Discussion    70.)     Filter,    treat   the   pre- 
cipitate by  (141)  and  reserve  the  filtrate  for  analysis  of 
Group  IV  (150). 

(141)  Separation  of  Arsenite.  —  Punch  a  hole  through 
the  filter  (140)  with  a  glass  rod  and  wash  the  precipitate 
into  a  graduate  cylinder  with  a  small  quantity  (15-20  cc.) 
of  water.     Add   2-3   cc.   of   NaOH    (approximately   6N.) 
and  fill  up  to  the  25  cc.  mark  with  water.     (See  Discus- 
sion 71.)     Transfer  the  contents  of  the  cylinder  to  a  beaker 
and  mix  thoroughly.     Filter  and  test  the  residue  for  arse- 
nite  (142)  and  treat  the  filtrate  by  (143). 

(142)  Detection  of  Arsenite.  —  Pour  repeatedly  through 
the  filter  (141)  a  5-cc.  portion  of  6N.  HC1  to  dissolve  the 
arsenite.     Saturate  the  solution  with  H2S.     A  yellow  pre- 
cipitate shows  the  presence  of  arsenite.     If  a  dark  pre- 
cipitate is  obtained  it  may  be  tested  for  arsenite  according 
to  the  method  outlined  for  arsenic.     (See  (21)  and  following.) 

(143)  Separation   of    Oxalate.  —  To    the    filtrate    (141) 
add  a  slight  excess  of  HCoHsOo  and  then  about  5  cc.  of 
NaC2H3O2  and  heat  to  boiling.     To  the  hot  solution  add 
an  excess  of  CaCl2  solution.     Mix  thoroughly  and  allow 
the  mixture  to  stand  for  two  to  three  minutes.     (See  Dis- 
cussion  72.)     The  formation  of  a  white  crystalline  pre- 
cipitate indicates  the  presence  of  oxalate.     Filter  and  con- 
firm the  oxalate  by  treating  the  precipitate  (CaC2O4)  by 
(144).     Treat  the  filtrate  by  (145). 

(144)  Detection  of  Oxalate.  —  Dissolve  the  precipitate 
(143)  by  pouring  5-6  cc.  of  6N.  H2SO4  repeatedly  through 
the  filter.     Warm  the  filtrate  and  add  a  drop  of  KMnO4 


104  ACIDS 

solution.     If    oxalate    is    present    the    permanganate    will 
be  decolorized. 

Optional  Method. — Instead  of  adding  KMnO4  transfer 
the  H2SO4  solution  to  a  flask  (Fig.  4)  and  boil  it,  at  the 
same  time  collecting  the  evolved  gas  in  lime-water.  If 
oxalate  is  present  it  will  decompose  with  the  formation 
of  CO2,  which  in  turn  will  react  with  the  Ca(OH)2  to  form 
insoluble  CaCO3.  (See  (130)  ). 

(145)  Separation  of  Arsenate  and  Phosphate. — To  the 
filtrate  (143)  add  uranyl  acetate  [UO2(C2H3O2)2]  in  slight 
excess.     Allow  the   mixture  to  stand  about  five  minutes 
and  filter.     (See  Discussion  73.)     Treat  the  precipitate  by 
(146)  and  the  filtrate  by  (147). 

(146)  Detection  of  Arsenate. — Dissolve  the  precipitate 
(145)    by  pouring  repeatedly  through  the  filter  a   lo-cc. 
portion  of  6N.  HC1;    heat  to  boiling  and  saturate  with 
H2S.     The  appearance  of  a  white,  finely  divided  precipi- 
tate, turning  yellow,  shows  the  presence  of  arsenate.     (See 
Discussion  17.)     Filter  and  treat  the  filtrate  by  (147). 

(147)  Detection    of    Phosphate. — If    the    presence    of 
phosphate  has  not  already  been  determined  (72),  evaporate 
to  dryness  the  filtrate   (146)   from  which  all  arsenic  has 
been  removed,  dissolve  the  residue  in  5  cc.  of  i6N.  HNOs, 
and  heat  to  drive  off  the  brown  fumes.     Pour  the  resulting 
solution  into  about  three  times  its  volume  of  (NH4)2MoO4 
reagent.     The  formation  of  a  yellow  crystalline  precipitate 
shows  the  presence  of  phosphate.     (See  Discussion  74.) 

(148)  Detection  of  Chromate. — Heat  the  filtrate  (145) 
to  boiling,  and  add  to  the  hot  solution  5  cc.  of  NH4C2H3O2 
and  an  excess  of  BaCl2  solution.     The  formation  of  a  yellow 
crystalline   precipitate   shows   the   presence   of   chromate. 
Unless  the  precipitate  is  yellow  and  crystalline  the  con- 
firmatory test  should  be  made.     (See  (63).) 

(149)  Detection  of  Tartrate.— To  the  filtrate  (148)  add 
about  10  cc.  of  6N.  H2SO4  and  filter  off  the  precipitated 
BaSO4.     Transfer   the   filtrate   to   a   porcelain   dish    and 
evaporate  to  5-10  cc.     The  presence  of  tartrate  is  indicated 


DISCUSSION  (70-75)  105 

by  the  ability  of  the  resulting  solution  to  decolorize  a 
drop  of  KMnCU  solution.     (See  Discussion  75.) 

DISCUSSION 

70.  The  presence  of  enough  silver  ions  to  p -capitate 
all  the  acids  of  Group   III  is  shown  by  the   'nmnentary 
formation  of  the  dark  Ag2O  precipitate  during  tae  addition 
of  NaOH.     The  Ag2O  dissolves  on  shaking,   as  long  as 
excess  acid  is  present  and  becomes  permanent  oaly  when 
all  acid  has  been  neutralized.     Its  failure  to  redissolve, 
therefore,  indicates  that  the  neutral  point  has  been  reached. 
If,  owing  to  the  precipitation  of  dark-colored  compounds 
of  metals  other  than  silver,  the  end-point  is  difficult  to 
detect,  it  may  be  ascertained  by  filtering  a  sir -all  portion 
and  adding  a  drop  of  phenolphthalein  and  thei  a  drop  of 
methyl  orange.     The  solution  should  remain  colorless  on 
the   addition    of   phenolphthalein    and    change   to   yellow 
when    methyl   orange   is   added.     If   the   solution   should 
change  to  red  on  the  addition  of  phenolphthaleia  too  much 
NaOH  has  been  added.     If  this  has  been  done  the  solution 
should  be  acidified  with  HNO3  and  NaOH  agaia  added. 

The  precipitation  should  be  made  in  lukewarm  solution, 
since  AgBO2  is  somewhat  insoluble  in  cold  neutral  solution. 

71.  Owing  to  the  formation  of  complex  ions  v/it/i  NaOH. 
and  the  slightly  soluble  Ag2O,  all  the  silver  salts  except 
arsenite  are  dissolved  in  an  excess  of  NaOH.     AgsAsOa 
remains  undissolved,  even  in  concentrations  as  high  as  I.5N. 
NaOH  solution,  while  as  much  as  250  mgs.  of  the  silver  salts 
of  the  other  acids  readily  dissolve  in  O.5N.  NaOH  solution. 

72.  Complete  removal  of  the  oxalate  is  essential,  since 
its  presence  would  later  interfere  with  the  detection  of 
tartrate  by  its  reducing  action  on  uranium.     Its  complete 
removal  is  assured  only  by  precipitation  from  hot  solution 
and  subsequent  digestion. 

73.  Arsenate  and  phosphate  form  insoluble  TJO2HAsO4 
and  UO2HPO4  respectively,  in  HC2H3O2  solution.     Both 
compounds  when  present  in  small  amounts  tend  to  pre- 


106  ACIDS 

cipitate  in  colloidal  or  finely  divided  form.  This  is  pre- 
vented somewhat  by  precipitation  from  hot  solution  or 
allowing  the  precipitate  to  stand.  Should  the  precipitate 
tend  to  run  through  the  filter  it  may  be  collected  and 
filtered  by  mixing  with  a  small  quantity  of  asbestos  fiber. 

74.  Both  phosphate  and  arsenate  react  with  (NH^oMoC^ 
reagent    to    form    insoluble    compounds  which    are    very 
similar  in   character  and    color.      The   complete   removal 
of  arsenate  is  therefore  necessary  before  testing  for  phos- 
phate.       Phosphate     is     most     easily    precipitated     with 
(NH^MoOi  reagent  in  a  strong  HNOs  solution  and  at  a 
temperature  of  about  60°  C. 

75.  Uranyl    salts   are    reduced    to    the    green    uranous 
state  on  boiling  a  strongly  acid  solution  with  tartrate  or 
other  reducing  agents.     On  the  other  hand,  in  the  absence 
of  reducing  agents  uranyl  salts  remain  unchanged  in  boil- 
ing cone.  H2SO4.     Since  oxalate  and  tartrate  are  by  far 
the  most  commonly  met  of  all  organic  acids  capable  of 
reducing  the  uranyl   compounds,    the   appearance   of   the 
green  uranous  salt  on  boiling  the  H2SO4  solution  is  indica- 
tive of  the  presence  of  tartrate,  oxalate  having  been  pre- 
viously removed.     The  absence  of  the  green  uranous  salt, 
as  shown  by  failure  to  reduce  KMnC>4  solution,  may  be 
taken  as  conclusive  evidence  of  the  absence  of  tartrate. 

If  other  organic  matter  may  be  present  and  the  uranous 
salt  is  obtained  in  (149),  a  portion  of  the  original  material 
may  be  tested  for  tartrate  as  follows:  Make  the  some- 
what concentrated  solution  slightly  alkaline  with  NH4OH, 
add  CaCb  solution  in  excess,  and  allow  the  mixture  to 
stand  a  short  time.  If  a  precipitate  forms  filter  and  digest 
the  precipitate  with  cold  NaOH  solution.  Dilute  slightly, 
filter,  and  boil  the  filtrate.  If  a  precipitate  forms,  filter 
while  hot,  wash  the  precipitate  and  transfer  it  to  a  test- 
tube.  Add  a  drop  of  NH4OH  and  a  little  AgNO3  solution, 
and  warm.  The  presence  of  tartrate  is  shown  by  the 
formation  of  a  black  precipitate  or  silver  mirror  on  the 
walls  of  the  tube. 


OUTLINES  OF  ANALYSIS  (TABLE  XV) 


107 


TABLE  XV 

OUTLINE  FOR  THE  SYSTEMATIC  SEPARATION  AND  DETECTION  OF  GROUP  IV 
Ions  present  in  neutral  solution 


No. 

Reagent 

SO4= 

BO3= 

C2H3O2- 

NO3~ 

i 

2 

3 

4 
5 

HC1 
BaCl2 
HC1 

turmeric 

H2SO4 
CH5OH 
FeS04 
H2S04 

H2SO4 
BaSO4 

H3BO3 
Ba3(BO3)2 

(?) 
Brownish- 
red  color 

1 

HC2H3O2 
Ba(C2H3O2)2 

Ba(C2H302)2 
C-2ri5(_x2ri3O2 

I 

HNO3 

< 

FeSO4-NO 

ANALYSIS 
Group  IV 
=  C2H302-,  NO3-,  F- 

(150)  Detection  of  Sulphate. — Evaporate  the  filtrate 
(140)  to  15-20  cc.     Add  slowly,  and  with  frequent  stirring 
just  enough  6N.  HC1  to  precipitate  all  the  silver;    filter 
and  reject  the  precipitate.     To  the  clear  filtrate  add  BaC^ 
solution  in  excess.     The  formation  of  a  white  crystalline 
precipitate   shows   the   presence   of   sulphate.      (See   Dis- 
cussion 76.)     Treat  the  filtrate  by  (151). 

(151)  Detection  of  Borate. — To  just  5  cc.  of  the  filtrate 
(150)  add  5  cc.  of  I2N.  HC1  and  2  drops  of  turmeric  solu- 
tion.    Allow  the  mixture  to  stand  ten  minutes.     If  borate 
is  present  the  solution  will  assume  a  brownish-red  color. 
(See  Discussion  77.) 

(152)  Detection   of  Acetate. — Test   a   portion  of  the 
original  substance  for  acetate  as  follows:    If  the  substance 
is  a  solid  add  to  a  small  quantity  on  a  watch  glass  a  few 
drops  of  cone.  H2SO4.     Mix  with  a  glass  rod  and  note 


108  ACIDS 

the  odor.  If  acetate  is  present  in  considerable  quantity 
the  odor  of  acetic  acid  will  be  apparent.  If  the  substance 
is  a  liquid  a  portion  of  the  slightly  alkaline  solution  should 
be  evaporated  to  dryness  and  tested  with  cone.  H2SO4 
as  directed  above.  (See  Discussion  78.) 

If  snail  quantities  are  suspected  the  following  test 
should  be  made:  To  a  small  portion  of  the  solid,  obtained 
by  evaporation  if  necessary,  add  about  0.5  cc.  of  alcohol 
and  i  cc  of  cone.  H2SO4.  Heat  gently  and  note  the  odor. 
The  ethylscetate  (C2H5C2H3O2)  formed  gives  a  character- 
istic pleasant  odor. 

(153)  Detection   of   Nitrate. — Test   a   portion   of   the 
original    substance    for    nitrate    as    follows:     Acidify    the 
solution  contained  in  a  test-tube  with  6N.  H2SO4,  and  add 
an  equa'   volume  of  FeSC>4  solution.     Pour  about  2  cc. 
of  cone.  H2SO4  slowly  down  the  sides  of  the  tube  so  that 
it  forms  a  layer  at  the  bottom  of  the  solution.     If  nitrate 
is  present  a  brown  ring  will  form  in  a  short  time  at  the 
juncture  oi  the  two  liquids.     (See  Discussion  79.) 

(154)  Detection  of   Fluoride. — In  the  absence  of  silica 
or  silicates  the  etching  test  may  be  made  as  follows:    Mix, 
with  the  aid  of    a    piece    of  wood,  about  I  gram  of  the 
powdered  material  in  a  lead  dish  or  platinum  crucible  with 
enough  cone.   H2SO4  to  form  a  thick  paste.     Cover  the 
dish  or  crucible  with  a  watch  glass  that  has  been  coated 
on  the  convex  side  with  a  thin  layer  of  paraffin  through 
which  characters  have  been  scratched.     Put  a  little  water 
in  the  watch  glass  to  prevent  melting  the  paraffin,  and 
warm  th(   dish  or  crucible  gently,  preferably  over  a  water 
bath,  for  some   time.     Remove  the  watch  glass,  melt  off 
the  pararfin  and  note  whether  the  parts  exposed  to  the 
action  oi   the  fumes  have  been  attacked.     If  fluoride  is 
present  the  glass  will  be  dissolved  off  or  etched  in  the 
places    that  were    exposed    to   the  fumes.      (See  Discus- 
sion 80.) 

In  the  presence  of  silica  or  silicates  it  is  evident  that  the 
etching  test  cannot  be  used.  If  the  material  can  be  decom- 


DISCUSSION  (76-80)  109 

posed  by  cone.  H2SO4  a  portion  of  the  powdered  material 
may  be  mixed  with  cone.  HoSCU  in  a  test-tube  and  warmed 
gently  while  a  drop  of  water  is  held  on  the  end  of  a  glass 
rod  in  the  vapors.  If  fluoride  is  present  the  drop  of  water 
will  become  turbid.  (See  Discussion  80.) 

If  the  material  cannot  be  decomposed  with  cone.  H2SO4 
it  may  be  fused  with  7-8  times  its  weight  of  a  riixture  of 
equal  parts  of  sodium  and  potassium  carbonate,  the  melt 
extracted  with  water,  filtered,  the  filtrate  acid  fied  with 
HC2H3O2  and  the  fluoride  precipitated  by  the  addition 
of  CaCl2  solution.  The  above  test  for  fluoride  may  then 
be  made  on  the  dried  precipitate. 

DISCUSSION 

76.  It  will  be  recalled  (see  Discussion  68)   th  it  one  of 
the  products  obtained  by  the  decomposition  of  Ag2S2O3 
is  H2SO4;  hence  if  thiosulphate  has  been  found,  a  portion 
of  the  original  solution  made  acid  with  HC1  should  be  used 
for  the  sulphate  test,  instead  of  solution  (150). 

77.  In  order  to  make  sure  of  the  presence  of  borate 
when  the  color  is  slight,  a  blank  test  should  be  made  for 
comparison.     In  order  to  be  able  to  estimate  th^  relative 
amount  present  the  student  should  make  comparison  tests 
with  known  amounts  of  borate.     In  either  case  ae  should 
take  care  to  make  all  tests  alike,  using  the  same  quantities 
of  HC1,   alcohol  and  turmeric,   since  the  shade  of  color 
depends  to  a  considerable  extent  upon  the  concentrations 
of  these  substances. 

78.  Although  acetate,  if  present  in  the  original  solu- 
tion, would  appear  in  the  filtrate  from  the  BaSO4  (150)  its 
detection  in  that  solution  would  necessitate  considerable 
evaporation  which  will  ordinarily  be  avoided  if  the  original 
substance  is  taken. 

79.  The  test  for  nitrate  is  very  delicate  and  accurate 
except  in  the  presence  of  certain  substances  which  cover 
up  the  ring  or  form  a  somewhat  similar  one.     Chromates 


110  ACIDS 

are  reduced  by  FeSO4  in  acid  solution  giving  a  green  color 
while  ferro-  and  ferricyanides  give  a  blue  precipitate  which 
makes  the  detection  of  the  brown  ring  more  difficult. 
Iodides  and  bromides  are  oxidized  by  the  cone.  H2SO4  and 
produce  a  ring  somewhat  similar  to  the  brown  of  the 
nitrate.  Chromates  may  be  removed  by  reduction  to 
chromic  salts  with  SO2  and  precipitation  with  NH4OH; 
the  others  may  be  removed  by  precipitation  from  slightly 
acid  solution  with  Ag2SO4  solution. 

80.  Most   fluorides  are   decomposed   by  cone.   H2SO4 
according  to  the  following  equation: 

CaF2  +H2S04->  CaS04  +H2F2 

The  reaction  proceeds  more  rapidly  if  the  mixture  is  heated, 
owing  to  the  more  rapid  removal  of  the  volatile  H2F2. 
The  H2F2  may  be  recognized  by  its  ability  to  etch  glass 
or  dissolve  silica  or  silicates.  In  either  case  volatile  silicon 
tetrafluoride  (SiF4)  is  formed.  When  this  is  brought  into 
contact  with  water  the  following  reaction  takes  place: 

SiF4  +4H2O->  2H2SiF6  +H4SiO4 


When  silicates  undecomposed  by  cone.  H2SO4  are  fused 
with  a  mixture  of  sodium  and  potassium  carbonates,  the 
silicon  reacts  to  form  soluble  Na2SiOs  and  any  fluoride 
present  forms  soluble  Na2F2. 


QUESTIONS  FOR  REVIEW  111 

QUESTIONS   FOR   REVIEW 

1.  What  ions  are  precipitated  by  AgNOs  in  acid  solution? 

2.  Why  should  the  solution  be  kept  cold  in  precipitating 
Group  I? 

3.  If   a   ferricyanide   is   present   what   other   acids  of 
Group  I  are  not  likely  to  be  present?     Why? 

4.  Although  it  is  possible  to  have  all  metals  in  the  same 
solution,  this  is  not  possible  in  the  case  of  acids.     Why? 

5.  Why  will  NaCl  solution  dissolve  all  the  cyanogen 
compounds  of  silver  but  have  no  effect  of  the  correspond- 
ing halides? 

6.  What   is   the   effect   of   boiling  Ag2S   with    HNO3? 
Write  equation. 

7.  What  is  the   principle  on   which   the  detection  of 
chloride,  bromide  and  iodide  is  based? 

8.  Why   is   it   necessary   to   remove   all   iodide   before 
testing    for    bromide?     Why    remove  all    bromide   before 
testing  for  chloride? 

9.  Given  a  solution  which  is  known  to  contain  no  other 
anions  than  those  given  below,  outline  a  method  of  analysis 
that  will  necessitate  no  unnecessary  steps; 

(a)  Ferrocyanide,  thiocyanate,  cyanide. 

(b)  Sulphide,  iodide,  chloride. 

(c)  Iodide,  bromide,  chloride. 

10.  In  driving  off  the  last  traces  of  bromide  before 
testing  for  chloride,  why  must  the  water  be  replaced  as  it 
evaporates? 

11.  Why  is  it  unlikely  that  Group   II   acids  will  be 
found  in  a  strongly  acid  solution? 

12.  What  is  the  precipitate  formed  in  the  test  for  sul- 
phite?    Write  equation. 

13.  Write  equations  showing  the  action  of  HNO2  on 
FeSO4. 

14.  If,  while  the  Group  II  anions  are  being  distilled  off 
a  white  precipitate  should  form  in  the  receiver  but  should 
redissolve  before  the  distillation  is  completed  what  conclu- 


112  ACIDS 

sions  couid  be  drawn  ?   Write  equations  showing  the  changes. 

15.  Why  should  Group  III  acids  be  precipitated  from 
a  warm  solution? 

1 6.  What  precautions  are  necessary  in  the  precipitation 
of  Group  1)1? 

17.  What  is  the  effect  of  adding  an  excess  of  NaOH  to 
the  silver  salts  of  Group  III? 

1 8.  Why  should  a  large  excess  of  NaOH  be  avoided  in  the 
separation  of  arsenite  from  the  remaining  Group  III  acids? 

19.  If  arsenate  is  found  why  must  it  be  completely 
removed  before  testing  for  phosphate? 

20.  What  is  the  confirmatory  test  for  oxalate?     Write 
equations  showing  all  reactions  involved. 

2 1 .  Whkt  is  the  black  precipitate  formed  on  the  addition 
of  NaOH  daring  the  precipitation  of  Group  III? 

22.  Why  must  an  excess  of  AgNOs  be  present  before 
precipitating  Group  III? 

23.  Wh?t  method  may  be  used  for  collecting  and  filter- 
ing a  colbidal  or  finely  divided  precipitate? 

24.  G:vcn  a  solution  which  is  known  to  contain  no  other 
anions  thar.  those  given  below  outline  a  method  of  analysis 
that  will  necessitate  no  unnecessary  steps: 

(a]  Nitrite,  arsenite,  arsenate. 

(b]  Oxalate,  chromate,  tartrate. 

(c]  Sulphite,  oxalate,  chromate. 

25.  What  effect  would  the  presence  of  shreds  of  filter 
paper  in  the  H2SO4  solution  have  on  the  test  for  tartrate? 

26.  Why  must  all  oxalate  be  removed  before  testing 
for  tartrate? 

27.  Could  Ba(NOs)2  be  used  in  the  place  of  HC1  and 
BaCU  in  the  test  for  sulphate? 

28.  What   acids   interfere  with   the   test   for   nitrate? 
How  ma>  they  be  removed? 

29.  Describe  the  test  for  fluoride. 

30.  If   z:nc  and   barium  were  found  in  an   unknown 
soluble  in  dilute  HNOs,  what  acids  would  it  be  unneces- 
sary to  test  for? 


APPENDIX 

I.    Preparation  of  Reagents 

Acids : 

Acetic,  6N. ;  Mix  350  cc.  of  glacial  acid  with  650  cc.  of 
water. 

Fluoboric;  dissolve  HsBOs  in  48  per  cent  H2F2,  using 
a  lead  or  platinum  dish,  until  a  test  portion  will  not 
give  a  precipitate  with  Pb(NO3)2  solution.  After 
cooling  add  an  equal  volume  of  alcohol  and  just 
enough  fluosilicic  acid  to  precipitate  any  sodium 
which  may  be  present  as  an  impurity. 

Fluosilicic;  pour  48  per  cent  H2F2  upon  sand  in  a  lead 
dish  or  wax  bottle  until  the  sand  is  barely  covered. 
Allow  the  mixture  to  stand  for  a  few  hours  or  until 
the  free  H2F2  has  entirely  disappeared.  Pour  off 
the  liquid  and  add  an  equal  volume  of  alcohol. 

Hydrochloric,  I2N.;  use  the  C.P.  acid  of  commerce 
(Sp.gr.  1.19), 

Hydrochloric,  6N.;  mix  equal  volumes  of  I2N.  HC1 
and  water. 

Hydrofluoric,  48  per  cent;  use  the  C.P.  acid  of  com- 
merce. 

Nitric,  i6N.;  use  the  C.P.  acid  of  commerce  (Sp.gr. 
1.42). 

Nitric,  6N.;  mix  380  cc.  of  i6N.  HNOs  with  620  cc. 
of  water. 

Perchloric,  6N.;  mix  650  cc.  of  the  60  per  cent  C.P. 
acid  with  350  cc.  of  water. 

Sulphuric,  96  per  cent;  use  the  C.P.  acid  of  commerce 
(Sp.gr.  1.84). 

Sulphuric,  6N. ;   mix  i  volume  of  the  96  per  cent  H2SC>4 
with  5  volumes  of  water. 
113 


114  APPENDIX 

Bases: 

» 

Ammonium   hydroxide,    I5N.;*.   use   the   C.P.    solution 

of  commerce  (Sp.gr.  0.90). 
Ammonium   hydroxide,  6N.;    mix  400  cc.  of  the  I5N. 

NH4OH  with  600  cc.  of  water. 
Calcium    hydroxide,    saturated    solution;     shake   5-10 

grams  of  C.P.   CaO  with   1000  cc.   of  water  until 

saturated,  and  filter. 
Sodium  hydroxide,  6N. ;    dissolve  250  grams  of    pure 

NaOH  in  enough  water  to  make  a  total  volume  of 

1000  cc. 

Salts: 

Ammonium  acetate,  3N.;  mix  equal  volumes  of  6N. 
HC2H3O2  and  6N.NH4OH,  or  dissolve  250  grams 
of  C.P.  NH4C2H3O2  in  enough  water  to  make  a 
total  volume  of  1000  cc. 

Ammonium  carbonate,  9N.;  dissolve  250  grams  of 
freshly  powdered  (NHU^COa  in  enough  cold  6N. 
NH4OH  to  make  1000  cc. 

Ammonium  chloride,  iN.;  dissolve  54  grams  of  NH4C1 
in  enough  water  to  make  a  total  volume  of  1000  cc. 

Ammonium  molybdate;  dissolve  75  grams  of  C.P. 
(NH4)2MoO4  in  500  cc.  of  water  and  pour  the  solu- 
tion slowly  into  500  cc.  of  6N.  HNOs.  If  a  pre- 
cipitate should  form,  shake  the  mixture  occasionally 
until  solution  is  complete. 

Ammonium  oxalate,  o.5N. ;  dissolve  35  grams  of 
(NH4)2C2O4-H2O  in  enough  water  to  make  a  total 
volume  of  1000  cc. 

Ammonium  polysulphide ;  saturate  500  cc.  of  6N. 
NH4OH  with  H2S  gas,  and  add  to  this  solution 
500  cc.  more  6N.  NH4OH,  50  cc.  of  6N.  NaOH  and 
25  grams  of  flowers  of  sulphur.  Digest  for  some 
hours  and  filter. 

Ammonium  sulphate,  iN.;  dissolve  66  grams  of 
(NH4)2SO4  in  1000  cc.  of  water. 


PREPARATION  FOR  REAGENTS  115 

Ammonium  sulphide;    saturate  500  cc.  of  6N.  NH4OH 

with    H2S   gas   and    then    add    500    cc.    more    6N. 

NH4OH. 
Barium  chloride,  iN. ;  dissolve  120  grams  of  BaCl2-2H2O 

in  1000  cc.  of  water. 
Bromine  water;   use  saturated  solution. 
Calcium  chloride,    iN.;    dissolve  150  grams  of  CaCl2- 

6H2O  in  1000  cc.  of  water. 
Calcium  sulphate;   use  saturated  solution. 
Cobalt  nitrate,  o.oiN. ;   dissolve  1.5  grams  of  Co(NO3)2- 

6H2O  in  1000  cc.  of  water. 
Dimethylglyoxime,    o.iN.;    dissolve    12   grams  of  the 

solid  in  1000  cc.  of  95  per  cent  alcohol. 
Ferric    chloride,    iN.;     dissolve    90    grams    of    FeCls- 

6H2O  in  1000  cc.  of  water. 
Ferrous  sulphate,  iN.;    dissolve  140  grams  of  FeSO4- 

yH2O  in  1000  cc.  of  o.6N.  IHbSOi  and  keep  in  con- 
tact with  iron  nails. 

Hydrogen  peroxide;   3  per  cent  solution. 
Lead  acetate,  o.2N. ;  dissolve  38  grams  of  Pb( 

3H2O  in  1000  cc.  of  water. 
Mercuric  chloride,  o.iN.;    dissolve  27  grams  of 

in  1000  cc.  of  water. 
Mercurous  nitrate,  o.iN. ;  dissolve  29  grams  of  HgNOs • 

2H2O  in  1000  cc.  of  water. 
Magnesia  mixture,  iN. ;   dissolve  100  grams  of  MgCb- 

6H2O  and  100  grams  of  NH4C1  in  water,  add  50  cc. 

of  I5N.  NH4OH  and  dilute  to  1000  cc. 
Potassium  cyanide,   iN.;    dissolve  65  grams  of  KCN 

in  1000  cc.  of  water. 
Potassium   dichromate,    iN.;    dissolve    147   grams   of 

K2Cr2O?  in  1000  cc.  of  water. 
Potassium  ferrocyanide,    iN.;    dissolve   105   grams  of 

K4Fe(CN)6-3H2O  in  1000  cc.  of  water. 
Potassium  iodide,  o.iN.;    dissolve  17  grams  of  KI  in 

1000  cc.  of  water. 


116  APPENDIX 

Potassium  nitrite,  3N.;    dissolve  250  grams  of  KNO2 

in  1000  cc.  of  water. 
Potassium  permanganate,  o.iN.;    dissolve  16  grams  of 

KMnC>4  in  1000  cc.  of  water. 
Potassium    thiocyanate,    iN.;     dissolve    97    grams    of 

KCNS  in  1000  cc.  of  water. 
Silver  nitrate,  O.5N.;    dissolve  85  grams  of  AgNOa  in 

1000  cc.  of  water. 
Sodium  carbonate,  3N.;  dissolve  159  grams  of  Na2COa 

in  1000  cc.  of  water. 
Sodium  chloride,   iN.;    dissolve  58  grams  of  NaCl  in 

1000  cc.  of  water  and  add  5  cc.  of  I2N.  HC1. 
Sodium     phosphate,     iN.;      dissolve     120     grams     of 

Na2HPO4-i2H2O  in  1000  cc.  of  water. 
Sodium  stannite;   add  NaOH,  drop  by  drop,  to  a  solu- 
tion of  SnCl2  until  the  precipitate  of  Sn(OH)2  is 

just   dissolved.     The   solution    must   be   kept   cold 

to  prevent  decomposition  and  oxidation  to  Na2SnC>3. 

Sodium  stannite  is  unstable  and  must  be  prepared 

as  needed. 
Stannous  chloride,  iN.;    dissolve  113  grams  of  SnCl2- 

2H2O  in  170  cc.  of  I2N.  HC1  and  dilute  to  1000  cc. 

Keep  in  bottles  containing  granulated  tin. 
Turmeric;    shake  an  excess  of  turmeric  powder  with 

95  per  cent  alcohol  and  filter. 
Uranium  acetate,  o.iN.;    dissolve  20  grams  of 

UO2(C2H302)2-2H20 

in  1000  cc.  of  water. 

Zinc  nitrate.   iN.;    dissolve  148  grams  of  Zn(NOs)2- 
6H2O  in  1000  cc.  of  water. 

II.    Test  Solutions 

Solutions  used  in  the  preliminary  experiments  should 
contain  equal  amounts  of  the  given  ions  per  cc.,  in  order 
that  the  student  may  learn  to  estimate  more  accurately 


TEST  SOLUTIONS 


117 


the  relative  amounts  of  the  constituents  in  an  unknown. 
The  solutions  given  below  when  dissolved  in  the  pro- 
portions stated  will  contain  10  mgs,  of  the  ion  per  cc. 
of  solution. 


Ion. 

Salt 

! 

Grams  per 
liter 

Ion. 

Salt 

Grams  per 
liter 

Ag 

AgNO3 

15-7 

Cr 

CrCl3 

30 

Hg 

HgNO3-2H2O 

15-     (*) 

Zn 

ZnCl2 

21 

Pb 

Pb(N03)2 

16.  (t) 

Mn 

Mn(NO3)2-6H2O 

52 

Hg 

HgCl2 

13-5 

Fe 

FeSO4-7H2O 

49      (II) 

Bi 

Bi(NO3)3-5H20 

23-     U) 

Co 

CoCl2-6H2O 

40 

Cu 

CuCl2-2H2O 

21 

Ni 

NiCl2-6H2O 

4i 

Cd 

CdCl2-2H2O 

19-5 

Ba 

BaCl2-2H2O 

18 

As 

NaAsO2 

17-3 

Sr 

SrCl2-6H2O 

30 

As 

Na2HAsO4-7H2O 

41 

Ca 

CaCl2-6H2O 

55 

Sb 

SbCl3 

19-       (§) 

Mg 

MgCl2-6H2O 

84 

Sn 

SnCl2-2H2O 

16.     (§) 

K 

KC1 

19 

Sn 

SnCl4-3H2O 

22.       (§) 

Na 

NaCl 

25 

Al 

A1C13 

49- 

*  Dissolve  in  0.6 N.  HNO33- 

f  Double  the  amount  for  Group  I  experiments. 

J  Dissolve  in  3N.HNO3. 

§  Dissolve  in  350  cc.  of  6N.  HCl  and  dilute  to  i  liter. 

f|  Dissolve  in  0.6 N.  H2SO4  and  keep  in  contact  with  iron  nails. 


118 


APPENDIX 


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rtrtaJcj        oj        rtrtrtrtaj 


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gpiqdinc     *  •»-- 1-  ^tttttt 

r.T.      L     O        .^M.-^.^H     Cy     Cu     Cu    C3 


03.-,    o3   o3   o3   o3   03   03   ^.-< 


water 
only. 


;  <*  *  s 

S-*?! 

s  w  8  .2 
•S  .s  .S  8 

^    «  .«  £ 

1111 

i? ;?  ?y 


a  =  soluble  in  acids. 
i  =  insoluble  in  water  and 


ydrolizes  completely  in  presence  of  w 
lightly  soluble  in  water,  soluble  in  aci 
soluble  in  water,  soluble  in  acids. 


wa 


A  LIST  OF  THE  MORE  COMMON  ELEMENTS 


119 


IV.  A  LIST  OF  THE  MORE  COMMON  ELEMENTS  WITH  THEIR  ATOMIC  WEIGHTS 


Name 

Symbol 

At.  Wt. 

Name 

Symbol 

At.  Wt. 

Aluminium 

Al 

27    I 

Lithium  . 

Li 

6    Q 

Antimony  
Arsenic  .... 

Sb 

As 

120.2 

7? 

Magnesium  .  .  . 
Manganese 

Mg 
Mn 

24-3 

C4.    Q 

Barium  

Ba 

208 

Mercury  

Hg 

2OO.6 

Boron  
Bromine  
Cadmium  

B 
Br 
Cd 

II 

79-9 
112.4 

Molybdenum 
Nickel  
Nitrogen    .... 

Mo 

Ni 
N 

96 

58.7 

Calcium  
Chlorine  .... 

Ca 

Cl 

40 

-it.    tr 

Oxygen  
Phosphorus  . 

0 
P 

16 

Chromium  
Cobalt  

Cr 
Co 

52 

58  9 

Potassium.  .  .  . 
Silicon    .... 

K 

Si 

39-1 
28  3 

Coooer  .  . 

Cu 

61   5 

Silver 

Ag 

IO7  8 

Fluorine  
Gold  
Hydrogen 

F 

Au 
H 

19 
197.2 
i  008 

Sodium  
Strontium.  .  .  . 
Sulphur 

Na 
Sr 

s 

23 
87.6 

-J2 

Iodine  
Iron          .... 

I 

Fe 

126.9 
55  8 

Tin  
Uranium 

Sn    . 

u 

II8.7 
2^8   2 

Lead  

Pb 

207.2 

Zinc  

Zn 

14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

RENEWALS  ONLY — TEL.  NO.  642-3405 

This  book  is  due  on  the  last  dace  stamped  below,  or 

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Renewed  books  are  subject  to  immediate  recall. 


maTWaa 

fef  AIM          1       K.                       *.     J. 

KK'DLD    MAR 

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Dpp'n    ,-, 

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"-V.   U 

MAV  24747$, 

ADD        rz  «nCX> 

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W>  1  3  1986      : 

RECEIVED 

JUL     Z  1986 

CIRCULATION  DEPT. 

LD2  1  A-60m.3/70                          IlniSSS^rfnSlStiiia 
(N5382slO)476-A-32                                     Berkeley 

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